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Indefinite Integrals

Learn indefinite integrals with examples. Solving indefinite integrals of trigonometric functions. Evaluating Definite Integral using FTC 2.

Indefinite Integrals

\(\int f(x)\;dx=F(x)\) means that \(F'(x)=f(x)\). We say that \( 'F'\)  is the indefinite integral of  \( 'f'\).

  • \(\int f(x)\;dx\) is the anti derivative function.

  • Indefinite integral represent an entire family of curves.

\(\int x^3\;dx=\dfrac {x^4}{4}+C\rightarrow\) each value of 'C' will give an anti-derivative.

\(\therefore \)  \(\dfrac {x^4}{4}+1,\;\dfrac {x^4}{4}-2,\;\;;\dfrac {x^4}{4}-6\) are all anti-derivatives of \(x^3\).

\(\int \limits_a^b f(x)dx=\int f(x)dx\Big]_a^b\)

  • \(\int_a^b\;f(x)\;dx\) is a value whereas \(\int f(x)dx\) is a function.

Consider 

\(I=\int\;x^2\;dx=\underbrace {\dfrac {x^3}{3}+C}_{function \;of \;x}\)

whereas \(\int\limits_1^2\;x^2\;dx=\dfrac {x^3}{3} \Bigg]_1^2=\dfrac {8}{3}-\dfrac {1}{3}=\underbrace{\dfrac {7}{3}}_{value}\)

Illustration Questions

If  \(\dfrac {d}{dx}\Big(-cos(e^x)+C\Big)=e^x\,sin\,e^x\) then \(\int e^x\,(sin\,e\,^x)\;dx\) will be

A \(-cos\,e^x+C\)

B \(\dfrac {x^2}{2}+C\)

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

D \(x^3+C\)

×

If \(\dfrac {d}{dx}\,f(x)=g(x)\), then \(\int g(x)\;dx=f(x)+C\)

\(\therefore \;\dfrac {d}{dx}\,(-cos\,e^x+C)=e^x\,sin\,e^x\), then \(\int e^x\,sin\,e^x\;dx=-cos\,e^x+C\)

Integration and differentiation are inverse process of each other.

If  \(\dfrac {d}{dx}\Big(-cos(e^x)+C\Big)=e^x\,sin\,e^x\) then \(\int e^x\,(sin\,e\,^x)\;dx\) will be

A

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

.

B

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

C

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

D

\(x^3+C\)

Option A is Correct

Rules of Indefinite Integrals

(1) \(\int c\,f(x)\;dx=c\int\,f(x)\;dx\)

(2) \(\int \Big(f(x)+g(x)\Big)\;dx=\int f(x)dx+\int g(x)\;dx\)

(3) \(\int x^n\;dx=\dfrac {x^n+1}{n+1}+C\) (for all constants \(n\))

(4) \(\int k\;dx=kx+C\) where \('k'\) is a constant

Illustration Questions

Find the general indefinite integral: \(I=\displaystyle\int\Bigg(2x^2+\sqrt x-\dfrac {1}{x\sqrt x}\Bigg)\;dx\)

A \(x^7+x^5-\sqrt x+C\)

B \(2x^2-\sqrt x+7x\sqrt x+C\)

C \(4x^3+x\sqrt x-\dfrac {1}{x}+C\)

D \(\dfrac {2x^3}{3}+\dfrac {2}{3}x^{3/2}+2x^{-1/2}+C\)

×

\(I=\int\,2x^2+\sqrt x-\dfrac {1}{x\sqrt x}\;dx\)

\(=\displaystyle\int\,2x^2\;dx+\int\sqrt x\;dx-\int\dfrac {1}{x\sqrt x}\;dx \)

\(=2\displaystyle\int\,x^2\;dx+\int x^{1/2}\;dx-\int\;x^{-3/2}\;dx\)

\(=2\dfrac {x^3}{3}+\dfrac {x^{3/2}}{3/2}-\dfrac {x^{-1/2}}{-1/2}+C\)    

   (  By using \(\int x^n\;dx=\dfrac {x^{n+1}}{n+1}+C\))

\(=\dfrac {2x^3}{3}+\dfrac {2}{3}\,x^{3/2}+2x^{-1/2}+C\)

Find the general indefinite integral: \(I=\displaystyle\int\Bigg(2x^2+\sqrt x-\dfrac {1}{x\sqrt x}\Bigg)\;dx\)

A

\(x^7+x^5-\sqrt x+C\)

.

B

\(2x^2-\sqrt x+7x\sqrt x+C\)

C

\(4x^3+x\sqrt x-\dfrac {1}{x}+C\)

D

\(\dfrac {2x^3}{3}+\dfrac {2}{3}x^{3/2}+2x^{-1/2}+C\)

Option D is Correct

Indefinite Integrals of Trigonometric Functions

(1) \(\displaystyle\int\,sin\,x\;dx=-cos\,x+C\)

(2) \(\displaystyle\int\,cos\,x\;dx=sin\,x+C\)

(3) \(\displaystyle\int\,sec^2\,x\;dx=tan\,x+C\)

(4) \(\displaystyle\int\,cosec^2\,x\;dx=-cot\,x+C\)

(5) \(\displaystyle\int\,sec\,x\;tan\,x\;dx=sec\,x+C\)

(6) \(\displaystyle\int\,cosec\,x\;cot\,x\;dx=-cosec\,x+C\)

All these can be verified by differentiating the right hand function.

Illustration Questions

Find the following general integral: \(I=\int2\,tan^2\,x\;dx\)

A \(2(tan\,x-x)+C\)

B \(2\,cos\,x-\dfrac {1}{x}+C\)

C \(2\,sec\,x+x+C\)

D \(2\,sin\,x+\dfrac {2}{x}+C\)

×

\(I=\int2\,tan^2\,x\;dx\)

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

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

\(=2\left [ \int sec^2\,x\;dx-\int 1\;dx \right]\)

\(=2\left [ tan\,x-x \right]+C\)

Find the following general integral: \(I=\int2\,tan^2\,x\;dx\)

A

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

.

B

\(2\,cos\,x-\dfrac {1}{x}+C\)

C

\(2\,sec\,x+x+C\)

D

\(2\,sin\,x+\dfrac {2}{x}+C\)

Option A is Correct

Evaluating Definite Integral using FTC 2 and formulas of Indefinite Integration

\(\displaystyle\int_a^b\,f(x)\;dx=F(b)-F(a)\)  where \(\displaystyle\int f(x)=F(x)+C\)

The integration constant 'C' has no role in the definite integral as it cancels out.

\(\displaystyle\int_a^b\,f(x)\;dx=F(x)+C\Bigg]_a^b\)

\(=\Big(F(b)+C\Big)-\Big(F(a)+C\Big)\)

\(=F(b)+C-F(a)-C\)

\(=F(b)-F(a)\)

Illustration Questions

Evaluate : \(\displaystyle\int_{-3}^1(2x^2-x+1)\;dx\)

A \(\dfrac {80}{3}\)

B \(14\)

C \(\dfrac {-17}{3}\)

D \(\dfrac {42}{5}\)

×

\(I=\displaystyle\int_{-3}^1(2x^2-x+1)\;dx\)

\(=\dfrac {2x^3}{3}-\dfrac {x^2}{2}+x\Bigg]_{-3}^1\)

\(=\left (\dfrac {2}{3}×1^3-\dfrac {1^2}{2}+1\right)\)\(-\left [2×\dfrac {(-3)^3}{3}-\dfrac {(-3)^2}{2}+(-3)\right]\)

\(=\left (\dfrac {2}{3}-\dfrac {1}{2}+1\right)- \left (2×(-9)-\dfrac {9}{2}-3\right)\)

\(=\dfrac {7}{6}-\left (\dfrac {-51}{2}\right) \)

\(=\dfrac {7}{6}+\dfrac {51}{2}\)

\(=\dfrac {160}{6}\)

\(=\dfrac {80}{3}\)

Evaluate : \(\displaystyle\int_{-3}^1(2x^2-x+1)\;dx\)

A

\(\dfrac {80}{3}\)

.

B

\(14\)

C

\(\dfrac {-17}{3}\)

D

\(\dfrac {42}{5}\)

Option A is Correct

Indefinite Integration

In some cases, it is required to split the numerator to solve the indefinite integration. 

e.g.  \(\displaystyle \int \dfrac{x^2+2x+1}{x+1}dx\) 

\(\Rightarrow \displaystyle\int \dfrac{(x+1)(x+1)}{x+1}dx\)

\(\Rightarrow \displaystyle\int (x+1)dx\)

\(\Rightarrow \dfrac{x^2}{2} +x+C\)

Illustration Questions

Evaluate:  \( \displaystyle\int \dfrac{sin\,2\,x}{sin\,x}dx\)

A \(2\,cos\,x\)

B \(2\,sin\,x+C\)

C \(cos\,x\)

D \(sin\,x+C\)

×

\(I=\displaystyle\int \dfrac{sin\,2\,x}{sin\,x}dx\)

\(\Rightarrow I=\displaystyle\int \dfrac{2\,sin\,x\,cos\,x}{sin\,x}dx\)

\(\Rightarrow I= \displaystyle\int 2\,cos\,x\,dx\)

\(= 2\displaystyle\int\,cos\,x\,dx\)

\(=2\,sin\,x+C\)

Evaluate:  \( \displaystyle\int \dfrac{sin\,2\,x}{sin\,x}dx\)

A

\(2\,cos\,x\)

.

B

\(2\,sin\,x+C\)

C

\(cos\,x\)

D

\(sin\,x+C\)

Option B is Correct

Mixed Integral Problems

Rules of Indefinite Integrals:

(1) \(\int c\,f(x)\;dx=c\int\,f(x)\;dx\)

(2) \(\int \Big(f(x)+g(x)\Big)\;dx=\int f(x)dx+\int g(x)\;dx\)

(3) \(\int x^n\;dx=\dfrac {x^n+1}{n+1}+C\) (for all constants \(n\))

(4) \(\int k\;dx=kx+C\) where \('k'\) is a constant

Indefinite Integrals of Trigonometric Functions

(1) \(\displaystyle\int\,sin\,x\;dx=-cos\,x+C\)

(2) \(\displaystyle\int\,cos\,x\;dx=sin\,x+C\)

(3) \(\displaystyle\int\,sec^2\,x\;dx=tan\,x+C\)

(4) \(\displaystyle\int\,cosec^2\,x\;dx=-cot\,x+C\)

(5) \(\displaystyle\int\,sec\,x\;tan\,x\;dx=sec\,x+C\)

(6) \(\displaystyle\int\,cosec\,x\;cot\,x\;dx=-cosec\,x+C\)

All these can be verified by differentiating the right hand function.

Illustration Questions

The general indefinite integral \(I=\displaystyle\int\;\left( \dfrac {x^2+1}{\sqrt x}+\dfrac {1}{2\,cos^2\,x} \right)\;dx\) equals

A \(\dfrac {5}{4}\sqrt x+7x^{{7}/{2}}-sin\,x+C\)

B \(\dfrac {7}{8}\,cos^2x+5x^{3/2}+x-C\)

C \(\dfrac {2}{5}\,x^{5/2}+2x^{1/2}+\dfrac {1}{2}\,tan\,x+C\)

D \(8\,sin\,x-\dfrac {1}{x\sqrt x}+x^2+C\)

×

\(I=\displaystyle\int\;\left( \dfrac {x^2+1}{\sqrt x}+\dfrac {1}{2\,cos^2\,x} \right)\;dx\)

\(=\displaystyle\int\;\left( \dfrac {x^2}{\sqrt x}+\dfrac {1}{\sqrt x}+\dfrac {1}{2\,cos^2\,x} \right)\;dx\)

\(=\displaystyle\int\; \dfrac {x^2}{\sqrt x}\;dx+ \int\;\dfrac {1}{\sqrt x}\;dx+ \int\; \dfrac {1}{2\,cos^2\,x} \;dx\)

\(=\displaystyle\int\; {x^{3/2}}\;dx+ \int\;{x^{-1/2}}\;dx+ \dfrac{1}{2}\int\; sec^2x \;dx\)

\(= \dfrac {x^{5/2}}{5/2}\;+ \dfrac {x^{1/2}}{1/2}\;+ \dfrac {1}{2}\,tan\,x+C\)

\(= \dfrac {2}{5}\;x^{5/2}\;+ 2\,x^{1/2}+ \dfrac {1}{2}\,tan\,x+C\)

The general indefinite integral \(I=\displaystyle\int\;\left( \dfrac {x^2+1}{\sqrt x}+\dfrac {1}{2\,cos^2\,x} \right)\;dx\) equals

A

\(\dfrac {5}{4}\sqrt x+7x^{{7}/{2}}-sin\,x+C\)

.

B

\(\dfrac {7}{8}\,cos^2x+5x^{3/2}+x-C\)

C

\(\dfrac {2}{5}\,x^{5/2}+2x^{1/2}+\dfrac {1}{2}\,tan\,x+C\)

D

\(8\,sin\,x-\dfrac {1}{x\sqrt x}+x^2+C\)

Option C is Correct

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