Chapter 7: Integrals

Ans:  We use the method of inspection as follows:

$\frac{d}{dx}\left( \cos 2x \right)=-2\sin 2x\Rightarrow -\frac{1}{2}\frac{d}{dx}\left( \cos 2x \right) $ 

$\therefore \sin 2x=\frac{d}{dx}\left( -\frac{1}{2}\cos 2x \right) $ 

Thus, the anti-derivative of sin $2x$ is $-\frac{1}{2}\cos 2x$.

Ans: We use the method of inspection as follows:

$\frac{d}{dx}\left( \sin 3x \right)=3\cos 3x\Rightarrow \frac{1}{3}\frac{d}{dx}\left( \sin 3x \right) $ 

$\therefore \cos 3x=\frac{d}{dx}\left( \frac{1}{3}\sin 3x \right) $ 

Thus, the anti - derivative of cos $3x$ is $\frac{1}{3}\sin 3x$.

Ans: We use the method of inspection as follows:

$\frac{d}{dx}\left( {{e}^{2x}} \right)\Rightarrow 2{{e}^{2x}}=\frac{1}{2}\frac{d}{dx}\left( {{e}^{2x}} \right) $ 

$\therefore {{e}^{2x}}=\frac{d}{dx}\left( \frac{1}{2}{{e}^{2x}} \right) $ 

Thus, the anti-derivative of ${{e}^{2x}}$ is $\frac{1}{2}{{e}^{2x}}$.

Ans: We use the method of inspection as follows:

$\frac{d}{dx}{{\left( ax+b \right)}^{3}}=3a{{\left( ax+b \right)}^{2}} $ 

$\Rightarrow {{\left( ax+b \right)}^{2}}=\frac{1}{3a}\frac{d}{dx}{{\left( ax+b \right)}^{3}} $ 

$\therefore {{\left( ax+b \right)}^{2}}=\frac{d}{dx}\left( \frac{1}{3a}{{\left( ax+b \right)}^{3}} \right) $ 

Thus, the anti-derivative of ${{\left( ax+b \right)}^{2}}$ is $\frac{1}{3a}{{\left( ax+b \right)}^{3}}$.

Ans: We use the method of inspection as follows:

$\frac{d}{dx}\left( -\frac{1}{2}\cos 2x-\frac{4}{3}{{e}^{3x}} \right)=\left( \sin 2x-4{{e}^{3x}} \right)$

Thus, the anti-derivative of $\left( \sin 2x-4{{e}^{3x}} \right)$ is $\left( -\frac{1}{2}\cos 2x-\frac{4}{3}{{e}^{3x}} \right)$.

Ans: 

$\int{\left( 4{{e}^{3x}}+1 \right)dx} $ 

$=4\int{{{e}^{3x}}dx}+\int{1}dx $ 

$=4\left( \frac{{{e}^{3x}}}{3} \right)+x+C $ 

$=\frac{4}{3}{{e}^{3x}}+x+C $

Ans: 

$\int{{{x}^{2}}\left( 1-\frac{1}{{{x}^{2}}} \right)dx} $ 

$=\int{\left( {{x}^{2}}-1 \right)dx} $ 

$=\frac{{{x}^{3}}}{3}-x+C $

Ans:

$\int{\left( a{{x}^{2}}+bx+c \right)}dx $ 

$=a\int{{{x}^{2}}dx+b\int{xdx+c\int{1.dx}}} $ 

$=a\left( \frac{{{x}^{3}}}{3} \right)+b\left( \frac{{{x}^{2}}}{2} \right)+cx+D $ 

$=\frac{a{{x}^{3}}}{3}+\frac{b{{x}^{2}}}{2}+cx+D $

Ans:  

$\int{\left( 2{{x}^{2}}+{{e}^{x}} \right)dx} $ 

$=2\int{{{x}^{2}}dx+\int{{{e}^{x}}dx}} $ 

$=2\left( \frac{{{x}^{3}}}{3} \right)+{{e}^{x}}+C $ 

$=\frac{2}{3}{{x}^{3}}+{{e}^{x}}+C $

Ans:

$\int{{{\left( \sqrt{x}-\frac{1}{\sqrt{x}} \right)}^{2}}}dx $ 

$=\int{\left( x+\frac{1}{x}-2 \right)dx} $ 

$=\int{xdx}+\int{\frac{1}{x}dx}-2\int{1.dx} $ 

$=\frac{{{x}^{2}}}{2}+\log \left| x \right|-2x+C $