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

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

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

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.

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

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

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

$$=F(b)-F(a)$$

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

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.

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

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.

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

Indefinite Integration

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

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

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

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

$$\Rightarrow \dfrac{x^2}{2} +x+C$$

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