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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.

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.

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

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.

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.

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 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 $$t=\sec x$$

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

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

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.

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.

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 $$II$$ .

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

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