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Integration Of Power Of Tan And Cot Functions Other Than Sin And Cos

Practice integrals using reduction formula & integration by parts formula, and evaluate definite integral which involves trigonometric functions.

Finding Integrals Using Given Reduction Formula

Reduction formula: 

  • It is the formula through which the higher powers of a function can be written in the lower powers of the same function.

Use of reduction formula

  • The main use of reduction formula is to find the integrals of higher powers of functions like \(\sin,\;\cos,\;\tan,\;\cot\), etc.
  • Each function has a different reduction formula.
  • And reduction formula for different functions can be derived easily.

Reduction formula for some functions are as follows:

Reduction formula for sin

\(\displaystyle\int\sin^nx\,dx=\dfrac{-1}{n}\cos x\,\sin^{n-1}x+\dfrac{n-1}{n}\int\sin^{n-2}x\,dx\)

Reduction formula for \(\cos x\)

\(\displaystyle\int\cos^nx\,dx=\dfrac{1}{n}\cos^{n-1} x\,\sin x+\dfrac{n-1}{n}\int\cos^{n-2}x\,dx\)

Reduction formula for \(\tan\)

\(\displaystyle\int\tan^nx\,dx=\dfrac{1}{n-1}\tan x-\int\tan^{n-2}x\,dx\)

Reduction formula for \(\cot\)

\(\displaystyle\int\cot^nx\,dx=\dfrac{-1}{n-1}\cot^{n-1} x-\int\cot^{n-2}x\,dx\)

How to use reduction formula?

  • To understand this, let us take can example.
  • We will find the integral of \(\sin^3x\) using reduction formula.

\(\displaystyle\int\sin^3x\,dx\)

Reduction formula for \(\sin x\)

\(\displaystyle\int\sin^nx\,dx=\dfrac{-1}{n}\cos x\,\sin^{n-1} x+\dfrac{n-1}{n}\int\sin^{n-2}x\,dx\)

Step 1: Let \(\displaystyle\int\sin^nx\,dx=I_n\) then \(I_n=\dfrac{-1}{n}\cos x\;\sin^{n-1}x+\dfrac{n-1}{n}I_{n-2}\)

put \(n=3\),

we need \(I_3=\dfrac{-1}{3}\cos x\,\sin^2x+\dfrac{2}{3}I_1\)

\(\displaystyle I_3=\dfrac{-1}{3}\cos x\,\sin^2x+\dfrac{2}{3}\int\sin x\,dx\)

  • Now, we know the integral of \(\sin x\).

So, \(\displaystyle\int\sin^3x\,dx=\dfrac{-1}{3}\cos x\,\sin^2x-\dfrac{2}{3}\cos x+C\)

  • We will add a constant of integration in the final answer.

Illustration Questions

Using the reduction formula \(\displaystyle\int (tan x)^n dx=\dfrac {(tan\,x)^{n-1}}{n-1}-\int (tan\;x)^{n-2}\;dx\) Evaluate \(\int (tan\,x)^4\;dx \)

A \(\dfrac {tan^3x}{3}-tan\,x+x+C\)

B \(\dfrac {tan^3x}{3}+x+C\)

C \(\dfrac {tan^2x}{2}-x^2+C\)

D \(x\,e^x+C\)

×

Let \(\int (tan\,x)^n\;dx=I_n\)

\(\therefore\,I_n=\dfrac {(tan\,x)^{n-1}}{n-1}-I_{n-2}\rightarrow\) given

We need

\(I_4=\dfrac {(tan\,x)^{3}}{3}-I_{2}\rightarrow\) put n = 4

\(I_2=tan\,x-I_{0}\rightarrow\) put n = 2

\(I_0=\int (tan\,x)^0\;dx=x+C\)

\(\therefore\;I_4=\dfrac {(tan\,x)^{3}}{3}-(tan\,x-x)+C\)

\(=\dfrac {(tan\,x)^{3}}{3}-tan\,x+x+C\)

Using the reduction formula \(\displaystyle\int (tan x)^n dx=\dfrac {(tan\,x)^{n-1}}{n-1}-\int (tan\;x)^{n-2}\;dx\) Evaluate \(\int (tan\,x)^4\;dx \)

A

\(\dfrac {tan^3x}{3}-tan\,x+x+C\)

.

B

\(\dfrac {tan^3x}{3}+x+C\)

C

\(\dfrac {tan^2x}{2}-x^2+C\)

D

\(x\,e^x+C\)

Option A is Correct

Evaluating Integrals involving Powers of Secant and Tangent

  • The integrals containing product of powers of secant and tangent are very complicated to evaluate.
  • But by using certain formulas, we can evaluate them easily.
  • There, arises a case for the integral like, \(\displaystyle I=\int(\sec x)^m(\tan x)^n\,dx\)

where \(m,\;n\) are integers \(m,\;n\geq0\)

Case: When the power of secant is even

When the power of secant is even, i.e., \(m\) is even, then by using the trigonometric identify, \(\sec^2x=1+\tan^2x\),

We can solve this easily.

For example: \(\int\sec^4x\;\tan^2x\,dx\)

Step 1: Separate \(\sec^2x\) and assume \(\tan x=t\).

We have done this, because the derivative of \(\tan x\) is \(\sec^2x\), and we can separate \(\sec^2x\) from \((\sec)^\text{even}\) which becomes \((\sec x)^2\;(\sec x)^\text{even –2}\)

This is based on the fact that even powers of \(\sec x\) can be expressed in terms of \(\tan x\) without radical sings.

\(\int\sec^2x\,\sec^2x\,\tan^2x\,dx\)

Let \(\tan x=t\)

Differentiating both sides w.r.t. \(x\)

\(\sec^2x\,dx=dt\)

Step 2: We will use above substitutions in the main integral,

\(\displaystyle\int\sec^2x\,\tan^2x\,\underbrace{\sec^2x\,dx}_{dt}\)

\(\int\sec^2x\,\tan^2x\,dt\)

Now, we will convert \(\sec^2x\) and \(\tan^2x\) in terms of \(t\) by using \(\sec^2x=1+\tan^2x\) and \(\tan x=t\)

So, \(\sec^2x=1+t^2\)

Note: Any even power of \(\sec x\) will be appropriate power of \(1+t^2\) after substitution.

\(\Rightarrow\;\int(1+t^2)t^2\,dt\)

\(=\int(t^2+t^4)\,dt\)

\(=\int t^2dt+\int t^4dt\)

\(=\dfrac{t^3}{3}+\dfrac{t^5}{5}+C\)

Step 3: \(=\dfrac{(\tan x)^3}{3}+\dfrac{(\tan x)^5}{5}+C\)

This is the required solution.

Illustration Questions

Evaluate \(I=\int tan^3\,x\;sec^2x\;dx\)

A \(\dfrac {tan^4\,x}{4}+C\)

B \(tan^2\,x\;sec^2x+C\)

C \(tan^3\,x\;sec\,x+C\)

D \(e^x+sin\,x+C\)

×

\(I=\int tan^3\,x\;sec^2x\;dx\)

Power of \(sec^\,x\) is even, therefore put \(tan^\,x=t\)

\(\Rightarrow sec^2x\;dx=dt\)

\(\therefore \;I=\int\,(tan^3x)(sec^2x\;dx)=\int t^3\;dt\)

\(=\dfrac {t^4}{4}+C\)

\(=\dfrac {tan^4\;x}{4}+C\)

Evaluate \(I=\int tan^3\,x\;sec^2x\;dx\)

A

\(\dfrac {tan^4\,x}{4}+C\)

.

B

\(tan^2\,x\;sec^2x+C\)

C

\(tan^3\,x\;sec\,x+C\)

D

\(e^x+sin\,x+C\)

Option A is Correct

Integrals of the Function of Powers of Secant and Tangent (when the power of Tangent Function is Odd)

  • We will learn to integrate the function of powers of secant and tangent.

Case: When the power of tangent function is odd

By using \(\tan^2x=\sec^2x-1\), we can solve this case.

For example : \(\int\tan^5x\,\sec^3x\,dx\)

Step 1: Separate \(\sec x\,\tan x\) and assume \(\sec x=t\).

We have done this, because the derivative of \(\sec x\) is \(\sec x\,\tan x\). This will remove \(\tan x\,\sec x\) from the integral and \(\tan^5x\) will become \(\tan^4x\) which can be written in terms of \(\sec x\) using \(\tan ^2x=\sec^2x-1\).

\(\int\sec x\,\tan x\,\tan^4x\,\sec^2x\,dx\)

Note : This is based on the fact that even powers of \(\tan x\) can be expressed in terms of \(\sec x\) without involving radical signs.

Let \(\sec x=t\)

Differentiating both sides w.r.t. \(x\)

\(\sec x\,\tan x\,dx=dt\)

Step 2: \(\displaystyle\int\sec^2x\,\tan^4x\;\underbrace{\sec x\,\tan x\,dx}_{dt}\)

\(\int\sec^2x\,\tan^4x\,dt\)

Now, express \(\sec^2x\) and \(\tan^4x\) in terms of \(t\) using \(\sec x=t\) and \(\tan^2x=\sec^2-1\)

\(\sec x=t\) and \(\tan ^4x=(\sec^2x-1)^2=\sec^2x-2\sec x+1=t^2-2t+1\)

So, \(\int t^2(t^2-2t+1)\,dt\)

\(=\int(t^2-2t^3+t^2)\,dt\)

\(=\int t^2dt-2\,\int t^3dt+\int t^2dt\)

\(=\dfrac{t^5}{5}-\dfrac{2t^5}{4}+\dfrac{t^3}{3}+C\)

Step 3: Put \(t=\sec x\)

\(\displaystyle\int\tan^5x\,\sec^3x\,dx=\dfrac{(\sec x)^5}{5}-\dfrac{(\sec x)^4}{4}+\dfrac{(\sec x)^3}{3}+C\)

Illustration Questions

Evaluate \(I=\int tan^3\,x\;sec\,x\;dx\)

A \(\dfrac {sec^2\,x}{2}-tan\,x+C\)

B \(x\,e^{x^2}+cos\,x+C\)

C \(\dfrac {sec^3\,x}{3}-sec\,x+C\)

D \((2x+1)\,cot\,x+C\)

×

\(I=\int tan^3\,x\;sec\,x\;dx\)

Power of \(tan^\,x\) is odd

\(\therefore\) Put \(sec^\,x=t\)

\(\Rightarrow sec^\,x\;tan\,x\;dx=dt\)

\(\therefore \;I=\int\,(tan^2x)×(tan\,x\;sec\,x)\;dx\)

\(=\int (t^2-1)\;dt \)      \((tan^2\,x=sec^2\,x-1 =t^2-1)\)

\(=\dfrac {t^3}{3}-t+C\)

\(=\dfrac {sec^3\;x}{3}-sec\,x+C\)

Evaluate \(I=\int tan^3\,x\;sec\,x\;dx\)

A

\(\dfrac {sec^2\,x}{2}-tan\,x+C\)

.

B

\(x\,e^{x^2}+cos\,x+C\)

C

\(\dfrac {sec^3\,x}{3}-sec\,x+C\)

D

\((2x+1)\,cot\,x+C\)

Option C is Correct

Definite Integrals of Trigonometric Functions

  • The definite integral of trigonometric function can be evaluated by making appropriate substitution and by changing the limit according to new variable.

For example : \(\int\limits^b_a(\sin x)^m\,(\cos x)^n\,dx\)

  • Here, we put \(\sin x=t\) or \(\cos x=t\) accordingly, as \(n\) is odd or \(m\) is odd.

Substitution for secant function having even powers

  • For the even powers of secant function, we substitute \(\tan x=t\), because the derivative of \(\tan x\) is \(\sec^2x\).

So, it reduces the power of secant function by 2 and makes it easy to integrate.

  • To understand this, let us take an example-

\(\int\limits^{\pi/4}_0\sec^6x\,dx\)

Step 1: Now, \(\tan x=t\)

Differentiating both sides with respect to \(x.\)

\(\sec^2x\,dx=dt\)

Step 2: Change the limits.

when \(x=0\) then \(t=0\)

when \(x=\dfrac{\pi}{4}\) then \(t=1\)

Step 3: \(\displaystyle\int\limits^1_0\sec^4x\,\underbrace{\sec^2x\,dx}_{dt}=\int\limits^1_0\sec^4x\,dt\)

We know even powers of \(\sec x\) can be written in terms of \(\tan x\).

Here, \(\sec^4x\) can be written in terms of \(t\) using \(1+\tan^2x=\sec^2x\)

\(\Rightarrow\;1+t^2=sec^2x\)

\(\displaystyle=\int\limits^1_0(1+t^2)^2dt=\int\limits^1_0(t^4+2t^2+1)\,dt=\left[\dfrac{t^5}{5}+\dfrac{2t^3}{3}+t\right]^1_0=\dfrac{1}{5}+\dfrac{2}{3}+1=\dfrac{28}{15}\)

This is the required solution.

Illustration Questions

Evaluate \(I=\int\limits_0^{\pi/4}\,(sec^4\,x)\;dx\)

A \(\dfrac {\pi}{2}\)

B \(\dfrac {3}{4}\)

C \(\dfrac {4}{3}\)

D \(\dfrac {\pi}{4}\)

×

\(I=\int\limits_0^{\pi/4}\,(sec^4\,x)\;dx\)

\(=\int\limits_0^{\pi/4}\,(sec^2\,x)\;(sec^2\,x)\;dx\)    (Separate \(sec^2\,x\))

Put  \(tan\,x=t \Rightarrow sec^2\,x\;dx=dt\)

when \(x=\dfrac {\pi}{4} \rightarrow t=1\)

\(x=0\rightarrow t=0\)

\(\therefore \;I=\int\limits_0^1(t^2+1)\;dt\)            \((sec^2\,x=1+tan^2\,x=1+t^2)\)

\(=\dfrac {t^3}{3}+t \Bigg]_0^1\)

\(=\Big(\dfrac {1}{3}+1\Big)-0\)

\(=\dfrac {4}{3}\)

Evaluate \(I=\int\limits_0^{\pi/4}\,(sec^4\,x)\;dx\)

A

\(\dfrac {\pi}{2}\)

.

B

\(\dfrac {3}{4}\)

C

\(\dfrac {4}{3}\)

D

\(\dfrac {\pi}{4}\)

Option C is Correct

Integrating Power of Tangents

  • We will integrate the powers of tangents by making appropriate substitutions and using formulas.
  • Let us consider on example:

\(\int\tan^nx\,dx\)

Step 1: Separate \(\tan^2x\,dx\)

\(=\int\tan^{n-2}x\cdot\tan^2x\,dx\)

Step 2: use \(\sec^2x-1=\tan^2x\)

\(=\int\tan^{n-2}x\cdot(\sec^2x-1)\,dx\)

\(=\int\left(\tan^{n-2}x\,\sec^2x-\tan^{n-2}x\right)\,dx\)

\(=\int\tan^{n-2}x\,\sec^2x\,dx-\int\tan^{n-2}x\,dx\)

For the first term, assume \(\tan x=t\)

Differentiate it with respect to \(x\)

\(\sec^2x\,dx=dt\)

Here, \(\sec^2x\,dx\) will vanish and only terms of \(\tan x\) will remain, which can be integrated easily.

For the second term, repeat the steps 1 and 2, i.e., separate \(\tan^2x\) and put \(\tan^2x=\sec^2x-1\).

Repeat this process until integrable form is obtained.

And at last we get the required solution.

For example : \(\int\tan^3x\,dx\)

Step 1: Separate \(\tan^2x\)

\(=\int\tan x\,\tan^2x\,dx\)

Step 2: Use \(\tan^2x=\sec^2x-1\)

\(=\int\tan x(\sec^2x-1)\,dx\)

\(=\int(\tan x\,\sec^2x-\tan x)\,dx\)

\(=\int\tan x\,\sec^2x\,dx-\int\tan x\,dx\)

For first term, put \(\tan x=t\)

Differentiating both sides with respect to \(x\)

\(\sec^2\,dx=dt\)

\(\displaystyle=\int \underbrace{\tan x}_{t}\;\underbrace{\sec^2x\,dx}_{dt}-\int\tan x\,dx\)

\(=\int t\,dt-\int \tan x\,dx\)

\(\Rightarrow\;\dfrac{t^2}{2}-\ell n|\sec x|+C\)

\(=\dfrac{\tan^2x}{2}-\ell n|\sec|+C\)

This is the required solution.

Illustration Questions

Evaluate \(I=\int tan^4x\;dx\)

A \(\dfrac {tan^3\,x}{3}-tan\,x+x+C\)

B \(sin^2\,x-3sin\,x\;cos\,x+C\)

C \(\dfrac {tan^2\,x}{2}-tan\,x+C\)

D \(x\,e^{2x}+\ell n\,x+C\)

×

\(I=\int tan^4x\;dx= \int tan^2x\;(tan^2\,x)\;dx\)

\(= \int tan^2x\;(sec^2\,x-1)\;dx\)

\(= \int \underbrace {tan^2x\;sec^2x\;dx}_{I_1}- \int \underbrace {tan^2x\;dx}_{I_2} \)

\(I_1=\int tan^2\,x\;sec^2\,x\;dx\)

Put \(tan\,x=t \Rightarrow sec^2\,x\;dx=dt\)

\(=\int t^2\;dt=\dfrac {t^3}{3}+C_1=\dfrac {tan^3\,x}{3}+C_1\)

\(I_2=\int tan^2\,x\;dx =\int sec^2\,x-1\;dx\)

\(=tan\,x-x+C_2\)

\(\therefore \;I=I_1-I_2=\dfrac {tan^3\,x}{3}-tan\,x+x+C\)

Evaluate \(I=\int tan^4x\;dx\)

A

\(\dfrac {tan^3\,x}{3}-tan\,x+x+C\)

.

B

\(sin^2\,x-3sin\,x\;cos\,x+C\)

C

\(\dfrac {tan^2\,x}{2}-tan\,x+C\)

D

\(x\,e^{2x}+\ell n\,x+C\)

Option A is Correct

Integrals involving Powers of Cosecant and Cotangent

  • The integrals containing product of powers of cosecant and cotangent are very complicated to evaluate.
  • But by using certain formulas, we can evaluate them easily.
  • There arises a case for the integral like,

\(I=\int(\cot_x)^m\;(\text{cosec}\,x)^ndx\) where \(m,\;n\) are integers and \(m,\,n\geq0\).

Case : When the power of cosecant is even

When the power of cosecant is even, then by using \(\text{cosec}^2x=1+\cot^2x\),

we can solve this easily.

For example : \(\int\cot x\,\text{cosec}^4x\,dx\)

Step 1: Separate \(\text{cosec}^2x\) and assume \(\cot x=t\)

We have done this, because the derivative of \(\cot x\) is \(\text{cosec}^2x\).

This will reduce the higher even power of \(\text{cosec}\,x\) to smaller even power.

And \(\text{cosec}\,x\) can be easily written in terms of \(\cot x\) by using \(\text{cosec}^2x=1+\cot^2x\).

\(\int\cot x\,\text{cosec}^2x\;\text{cosec}^2x\,dx\)

Let \(\cot x=t\)

Differentiating both sides with respect to \(x.\)

\(\text{cosec}^2x\,dx=dt\)

Step 2: We will use the above substitutions in the main integral.

\(\int\underbrace{\cot x}_{t}\;\text{cosec}^2x\;\underbrace{\text{cosec}^2x\,dx}_{dt}\)

\(\int t\,\text{cosec}^2x\,dt\)

Convert \(\text{cosec}^2x\) in terms of \(t.\)

\(\text{cosec}^2x=1+\cot^2x\)

\(=1+t^2\)

\(=\int t(1+t^2)\,dt\)

\(=\int(t+t^3)\,dt\)

\(=\int t\,dt+\int t^3dt\)

\(=\dfrac{t^2}{2}+\dfrac{t^4}{4}+C\)

Step 3: \(=\dfrac{(\cot x)^2}{2}+\dfrac{(\cot x)^4}{4}+C\)

This is the required solution.

Illustration Questions

Evaluate \(I=\int cot^3\,x\;cosec^2\,x\;dx\)

A \(\dfrac {-cot^4\,x}{4}+C\)

B \(\dfrac {-cot^3\,x}{3}+C\)

C \(-cot^2\,x+cosec\,x+C\)

D \(x^2\,\ell n\,x+C\)

×

\(I=\int cot^3\,x\;cosec^2\,x\;dx\)

Put \(cot\,x=t\Rightarrow -cosec^2\,x\;dx=dt\)

\(\therefore \;I=\int t^3\,×(-dt)=\dfrac {-t^4}{4}+C=\dfrac {-cot^4\,x}{4}+C\)

Evaluate \(I=\int cot^3\,x\;cosec^2\,x\;dx\)

A

\(\dfrac {-cot^4\,x}{4}+C\)

.

B

\(\dfrac {-cot^3\,x}{3}+C\)

C

\(-cot^2\,x+cosec\,x+C\)

D

\(x^2\,\ell n\,x+C\)

Option A is Correct

Evaluating Integrals involving Powers of Cosecant and Cotangent by trigonometric substitutions

  • Here, we will learn to evaluate the integrals involving powers of cosecant and cotangent functions.

Case : When the power of cotangent is odd

When the power of cotangent is odd, then by using \(\text{cosec}^2x=1+\cot^2x\),

We can solve this easily.

For example : \(\int \cot^5x\,\text{cosec}^5x\,dx\)

Step 1: Separate \(\text{cosec}\,x\,\cot x\) and assume \(\text{cosec}\,x=t\)

We have done this, because the derivative of \(\text{cosec}\,x\) is \(\text{cosec}\,x\;\cot x\), this will make \(\cot^5x\) even. And \(\cot^4x\) can be written in terms of \(\text{cosec}\,x\) by using \(\text{cosec}^2x=1+\cot^2x\)

\(\int\cot x\;\text{cosec}\,x\;\cot^4x\;\text{cosec}^4 x\,dx\)

Let \(\text{cosec}\,x=t\)

Differentiating both sides w.r.t. \(x\)

\(\text{cosec}\,x\;\cot x\,dx=-dt\)

Step 2: We will use the above substitutions in the main integral.

\(\int\cot^4x\;\underbrace{\text{cosec}^4x}_{t^4}\;\cot x\;\underbrace{\text{cosec}\,x\,dx}_{-dt}\)

Convert \(\cot^4x\) in terms of \(t\)

\(\text{cosec}^2x=1+\cot^2x\)

\(t^2=1+\cot^2x\)

\(\cot^2x=t^2-1\)

\(\cot^4x=(t^2-1)^2\)

\(=-\int(t^2-1)^2\,t^4\,dt\)

\(=-\int(t^4-2t^2+1)\,t^4\,dt\)

\(=-\int(t^8-2t^6+t^4)\,dt\)

\(=-\left[\int t^8dt-2\int t^6dt+\int t^4dt\right]\)

\(=-\left(\dfrac{t^9}{9}-\dfrac{2t^7}{7}+\dfrac{t^5}{5}\right)+C\)

Step 3: \(=-\dfrac{(\text{cosec}\,x)^9}{9}+\dfrac{2(\text{cosec}\,x)^7}{7}-\dfrac{(\text{cosec}\,x)^5}{5}+C\)

This is the required solution.

Illustration Questions

Evaluate \(I=\int cot^3\,x\;cosec^3x\;dx\)

A \(\dfrac {cosec^3\,x}{3} -\dfrac {cosec^5\,x}{5}+C\)

B \(cot^2\,x-tan^2\,x+C\)

C \(\dfrac {cosec^4\,x}{4} -cosec^2\,x+C\)

D \(x^2\,e^{2x}+\ell n\,x+C\)

×

\(I=\int cot^3\,x\;cosec^3x\;dx\)

\( \;I=\int cot^2\,x×cosec^2\,x×(cosec\,x\;cot\,x)\;dx\)

Put \(cosec\,x=t\Rightarrow -cosec\,x\;cot\,x\;dx=dt\)

 

\(\because\:=\int (t^2-1)\,t^2\;(-dt)\)

\(cot^2x=cosec^2x-1=t^2-1\))

\(=\int (t^2-t^4)\,dt\)

\(= \dfrac {t^3}{3}-\dfrac {t^5}{5}+C\)

\(=\dfrac {cosec^3\,x}{3} -\dfrac {cosec^5\,x}{5}+C\)

Evaluate \(I=\int cot^3\,x\;cosec^3x\;dx\)

A

\(\dfrac {cosec^3\,x}{3} -\dfrac {cosec^5\,x}{5}+C\)

.

B

\(cot^2\,x-tan^2\,x+C\)

C

\(\dfrac {cosec^4\,x}{4} -cosec^2\,x+C\)

D

\(x^2\,e^{2x}+\ell n\,x+C\)

Option A is Correct

Indefinite Integration by Parts

If we combine the formula for integration by parts with fundamentals of calculus \(II\) .

\(\Rightarrow \displaystyle\int f(x)\,g'(x)\,dx = f(x) \,g(x) - \displaystyle\int g(x)\,f'(x)\,dx\)

Illustration Questions

Evaluate \(I= \displaystyle\int\sqrt x \,\ell n\,x\,dx\) 

A \(\dfrac{2}{3} \,x^{3/_2}\,\ell n\,x - \dfrac{4}{9} \,x^{3/_2} +C\)

B \(\dfrac{3}{2} \,x^{3/_2}\,\ell n\,x - \dfrac{2}{9} \,x^{3/_2} +C\)

C \(\dfrac{4}{3} \,x^{4/_3}\,\ell n\,x -(\ell n \,x)^{3/_2} +C\)

D \(\dfrac{3}{2} \,x^{2/_3}\,\ell n\,x +C\)

×

\( \displaystyle\int f(x)\,g'(x)\,dx = f(x) \,g(x) - \displaystyle\int g(x)\,f'(x)\,dx\)

\(\Rightarrow g'(x) = x^{1/_2}\)

\(\Rightarrow g(x) = \displaystyle \int x^{1/_2}\,dx\)

\(=\dfrac{x^{3/_2}}{\dfrac{3}{2}} = \dfrac{2}{3} \,x^{3/_2}\)

\(I = \ell n \,x × \dfrac{2}{3} \,x^{3/_2} - \displaystyle \int \dfrac{1}{x} × \dfrac{2}{3}\,x^{3/_2} \,dx\)

\(I = \ell n \,x × \dfrac{2}{3} \,x^{3/_2} -\dfrac{2}{3} \displaystyle \int \,x^{1/_2} \,dx\)

\(= \dfrac{2}{3} \,x^{3/_2} \ell n \,x - \dfrac{2}{3} \dfrac{x^{3/_2}}{\dfrac{3}{2}} +C\)

\(= \dfrac{2}{3} \,x^{3/_2} \ell n\,x - \dfrac{4}{9} x^{3/_2} +C\)

Evaluate \(I= \displaystyle\int\sqrt x \,\ell n\,x\,dx\) 

A

\(\dfrac{2}{3} \,x^{3/_2}\,\ell n\,x - \dfrac{4}{9} \,x^{3/_2} +C\)

.

B

\(\dfrac{3}{2} \,x^{3/_2}\,\ell n\,x - \dfrac{2}{9} \,x^{3/_2} +C\)

C

\(\dfrac{4}{3} \,x^{4/_3}\,\ell n\,x -(\ell n \,x)^{3/_2} +C\)

D

\(\dfrac{3}{2} \,x^{2/_3}\,\ell n\,x +C\)

Option A is Correct

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