Chapter 7: Integrals

Ans:

Let  $ 1+{{x}^{2}}=\text{t} $ 

$ \therefore 2\text{xdx}=\text{dt} $ 

$ \Rightarrow \int{\dfrac{2x}{1+{{x}^{2}}}}dx=\int{\dfrac{1}{t}}dt $ 

$ =\log |t|+C $ 

$ =\log \left| 1+{{x}^{2}} \right|+C $ 

$ =\log \left( 1+{{x}^{2}} \right)+C $ 

where C is an arbitrary constant. 

Ans:

Let  $ \log x=t $ 

$ \therefore \dfrac{1}{x}dx=dt $ 

$ \Rightarrow \int{\dfrac{{{(\log |x|)}^{2}}}{x}}dx=\int{{{t}^{2}}}dt $ 

$ =\dfrac{{{t}^{3}}}{3}+C $ 

$ =\dfrac{{{(\log |x|)}^{3}}}{3}+C $ 

where C is an arbitrary constant. 

Ans:

The given function can be written as 

$ \dfrac{1}{x+x\log x}=\dfrac{1}{x\left( 1+\log x \right)} $ 

Let  $ 1+\log x=t $ 

$ \therefore \dfrac{1}{x}dx=dt $ 

$ \Rightarrow \int{\dfrac{1}{x(1+\log x)}}dx=\int_{t}^{1}{d}t $ 

$ =\log |t|+C $ 

$ =\log |1+\log x|+C $ 

where C is an arbitrary constant.

Ans:

Let  $ \cos x=t $ 

$ \therefore -\sin xdx=t $ 

$ \Rightarrow \int{\sin }x\cdot \sin (\cos x)dx=-\int{\sin }tdt $ 

$ =-[-\cos t]+C $ 

$ =\cos t+C $ 

$ =\cos (\cos x)+C $ 

where C is an arbitrary constant.

Ans:

The given function can be rewritten as

 $ \sin (ax+b)\cos (ax+b)=\dfrac{2\sin (ax+b)\cos (ax+b)}{2}=\dfrac{\sin 2(ax+b)}{2} $ 

let  $ 2(ax+b)=t $ 

 $ \therefore 2adx=dt $ 

 $ \Rightarrow \int{\dfrac{\sin 2(ax+b)}{2}}dx=\dfrac{1}{2}\int{\dfrac{\sin tdt}{2a}} $ 

 $ =\dfrac{1}{4a}[-\cos t]+C $ 

 $ =\dfrac{-1}{4a}\cos 2(ax+b)+C $ 

where C is an arbitrary constant.

Ans:

Let  $ ax+b=t $ 

\[\Rightarrow adx=dt\]

\[\therefore dx=\dfrac{1}{a}dt\]

\[\Rightarrow \int{{{(ax+b)}^{\dfrac{1}{2}}}}dx=\dfrac{1}{a}\int{{{t}^{\dfrac{1}{2}}}}dt\]

\[=\dfrac{1}{a}\left( \dfrac{{{t}^{\dfrac{1}{2}}}}{\dfrac{3}{2}} \right)+C\]

\[=\dfrac{2}{3a}{{(ax+b)}^{\dfrac{3}{2}}}+C\]

where C is an arbitrary constant.

Ans:

Let  $ x+2=t $ 

 $ \therefore dx=dt $ 

 $ \Rightarrow \int{x}\sqrt{x+2}dx=\int{(t-2)}\sqrt{t}dt $ 

 $ =\int{\left( {{t}^{\dfrac{3}{2}}}-2{{t}^{\dfrac{1}{2}}} \right)}dt\ $ \ $ =\int{{{t}^{\dfrac{3}{2}}}}dt-2\int{{{t}^{\dfrac{1}{2}}}}dt $ 

 $ =\dfrac{{{t}^{\dfrac{5}{2}}}}{\dfrac{5}{2}}-2\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $ 

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

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

where C is an arbitrary constant.

Ans:

Let  $ 1+2{{x}^{2}}=t $ 

\[\therefore 4\text{xdx}=\text{dt}\]

\[\Rightarrow \int{x}\sqrt{1+2{{x}^{2}}}dx=\int{\dfrac{\sqrt{t}}{4}}dt\]

\[=\dfrac{1}{4}\int{{{t}^{\dfrac{1}{2}}}}dt\]

\[=\dfrac{1}{4}\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C\]

\[=\dfrac{1}{6}{{\left( 1+2{{x}^{2}} \right)}^{\dfrac{3}{2}}}+C\]

where C is an arbitrary constant.

Ans:

Let  $ {{\text{x}}^{2}}+\text{x}+1=\text{t} $ 

 $ \therefore (2\text{x}+1)\text{dx}=\text{dt} $ 

 $ \int{(4x+2)}\sqrt{{{x}^{2}}+x+1}dx $ 

 $ =\int{2}\sqrt{t}dt $ 

 $ =2\int{\sqrt{t}}dt $ 

 $ =2\left( \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right)+C $ 

 $ =\dfrac{4}{3}{{\left( {{x}^{2}}+x+1 \right)}^{\dfrac{3}{2}}}+C $ 

where C is an arbitrary constant.

Ans:

The given function can be rewritten as

 $ \dfrac{1}{x-\sqrt{x}}=\dfrac{1}{\sqrt{x}\left( \sqrt{x}-1 \right)} $ 

Let $ \left( \sqrt{x}-1 \right)=t $ 

 $ \therefore \dfrac{1}{2\sqrt{x}}dx=dt $ 

 $ \Rightarrow \int{\dfrac{1}{\sqrt{x}(\sqrt{x}-1)}}dx=\int{\dfrac{2}{t}}dt $ 

 $ =2\log |t|+C $ 

 $ =2\log |\sqrt{x}-1|+C $ 

where C is an arbitrary constant.