Chapter 7: Integrals

Ans: Given $\dfrac{1}{x-{{x}^{3}}}$

So, $\dfrac{1}{x-{{x}^{3}}}=\dfrac{1}{x\left( 1-{{x}^{2}} \right)}$

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

Let $\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}=\dfrac{A}{x}+\dfrac{B}{\left( 1-x \right)}+\dfrac{C}{1+x}$                 …….(1)

$\Rightarrow 1=A\left( 1-{{x}^{2}} \right)+Bx\left( 1+x \right)+Cx\left( 1-x \right)$

$\Rightarrow 1=A-A{{x}^{2}}+Bx+B{{x}^{2}}+Cx-C{{x}^{2}}$

On equating the coefficients of ${{x}^{2}},x$ and constant term –

$-A+B-C=0$              

$B+C=0$

$A=1$

Thus, we get $A=1,B=\dfrac{1}{2}$ and $C=-\dfrac{1}{2}$

By equation 1-

$\dfrac{1}{x\left( 1-x \right)\left( 1+x \right)}=\dfrac{1}{x}+\dfrac{1}{2\left( 1-x \right)}-\dfrac{1}{2\left( 1+x \right)}$

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

$=\log x-\dfrac{1}{2}\log \left( 1-x \right)-\dfrac{1}{2}\log \left( 1+x \right)$

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

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

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

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

Ans: Given $\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}$

$=\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}\times \dfrac{\sqrt{x+a}-\sqrt{x+b}}{\sqrt{x+a}-\sqrt{x+b}}$

$=\dfrac{\sqrt{x+a}-\sqrt{x+b}}{\left( x+a \right)-\left( x+b \right)}$

$=\dfrac{\sqrt{x+a}-\sqrt{x+b}}{a-b}$

$\Rightarrow \int{\dfrac{1}{\sqrt{x+a}+\sqrt{x+b}}}dx=\dfrac{1}{a-b}\int{\left( \sqrt{x+a}-\sqrt{x+b} \right)}dx$

$=\dfrac{1}{a-b}\left[ \dfrac{{{\left( x+a \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}-\dfrac{{{\left( x+b \right)}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]$

$=\dfrac{2}{3\left( a-b \right)}\left[ {{\left( x+a \right)}^{\dfrac{3}{2}}}-{{\left( x+b \right)}^{\dfrac{3}{2}}} \right]+C$

Ans: Given

$\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}$

Let $x=\dfrac{a}{t}$

On differentiating-

$dx=-\dfrac{a}{{{t}^{2}}}dt$

$\Rightarrow \int{\dfrac{1}{x\sqrt{ax-{{x}^{2}}}}dx=\int{\dfrac{1}{\dfrac{a}{t}\sqrt{a.\dfrac{a}{t}-{{\left( \dfrac{a}{t} \right)}^{2}}}}}\left( -\dfrac{a}{{{t}^{2}}}dt \right)}$

$=-\int{\dfrac{1}{at}.\dfrac{1}{\sqrt{\dfrac{1}{t}-\dfrac{1}{{{t}^{2}}}}}}dt$

$=-\dfrac{1}{a}\int{\dfrac{1}{\sqrt{\dfrac{{{t}^{2}}}{t}-\dfrac{{{t}^{2}}}{{{t}^{2}}}}}}dt$

$=-\dfrac{1}{a}\int{\dfrac{1}{\sqrt{t-1}}}dt$

$=-\dfrac{1}{a}\left[ 2\sqrt{t-1} \right]+C$

Substituting value of t-

$=-\dfrac{1}{a}\left[ 2\sqrt{\dfrac{a}{x}-1} \right]+C$

$=-\dfrac{2}{a}\left[ \dfrac{\sqrt{a-x}}{\sqrt{x}} \right]+C$

$=-\dfrac{2}{a}\left[ \sqrt{\dfrac{a-x}{x}} \right]+C$

Ans:The given expression is, $\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}$. 

On multiplying and dividing by ${{x}^{-3}}$, the following can be obtained as shown below,

$\dfrac{{{x}^{-3}}}{{{x}^{2}}.{{x}^{-3}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}=\dfrac{{{x}^{-3}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{-3}{4}}}}{{{x}^{2}}-{{x}^{-3}}}=\dfrac{{{\left( {{x}^{4}}+1 \right)}^{\dfrac{-3}{4}}}}{{{x}^{5}}{{\left( {{x}^{4}} \right)}^{\dfrac{-3}{4}}}}=\dfrac{1}{{{x}^{5}}}{{\left( \dfrac{{{x}^{4}}+1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}=\dfrac{1}{{{x}^{5}}}{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}$ 

Now, consider $\dfrac{1}{{{x}^{4}}}=t$ 

$\therefore \dfrac{-4}{{{x}^{5}}}=\dfrac{dt}{dx}$

$\Rightarrow \dfrac{dx}{{{x}^{5}}}=\dfrac{-dt}{4}$

$\therefore \int{\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}dx}=\int{\dfrac{1}{{{x}^{5}}}{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{-3}{4}}}dx}=\dfrac{-1}{4}\int{{{\left( 1+t \right)}^{\dfrac{-3}{4}}}dt}$ 

$\Rightarrow \int{\dfrac{1}{{{x}^{2}}{{\left( {{x}^{4}}+1 \right)}^{\dfrac{3}{4}}}}dx}=\dfrac{-1}{4}\left[ \dfrac{{{\left( 1+t \right)}^{\dfrac{1}{4}}}}{\dfrac{1}{4}} \right]+C=-{{\left( 1+\dfrac{1}{{{x}^{4}}} \right)}^{\dfrac{1}{4}}}+C$, where $C$ is any arbitrary constant.

Ans: The given expression is, $\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}$. 

Observe the given hint and obtain as shown below,

$\therefore \dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}=\dfrac{1}{{{x}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{6}}} \right)}$ 

Consider $x={{t}^{6}}$ 

$\therefore x={{t}^{6}}\Rightarrow dx=6{{t}^{5}}$

$\therefore \int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}dx}=\int{\dfrac{1}{{{x}^{\dfrac{1}{3}}}\left( 1+{{x}^{\dfrac{1}{6}}} \right)}dx}=\int{\dfrac{6{{t}^{5}}}{{{t}^{2}}\left( 1+t \right)}dt}=6\int{\dfrac{{{t}^{3}}}{\left( 1+t \right)}dt}$ 

Now on dividing, we can obtain as shown below,

$\int{\dfrac{1}{{{x}^{\dfrac{1}{2}}}+{{x}^{\dfrac{1}{3}}}}dx}=6\int{\left\{ \left( {{t}^{2}}-t+1 \right)-\dfrac{1}{1+t} \right\}dt}$

$=6\int{\left[ \left( \dfrac{{{t}^{3}}}{3} \right)-\left( \dfrac{{{t}^{2}}}{2} \right)+t-\log \left| 1+t \right| \right]dt}$

$=2{{x}^{\dfrac{1}{2}}}-3{{x}^{\dfrac{1}{3}}}+6{{x}^{\dfrac{1}{6}}}-6\log \left( 1+{{x}^{\dfrac{1}{6}}} \right)+C$

$=2\sqrt{x}-3{{x}^{\dfrac{1}{3}}}+6{{x}^{\dfrac{1}{6}}}-6\log \left( 1+{{x}^{\dfrac{1}{6}}} \right)+C$, where $C$ is any arbitrary constant.

Ans: The given expression is, $\dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}$. 

Now consider it as shown below,

$\therefore \dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}=\dfrac{A}{\left( x+1 \right)}+\dfrac{Bx+C}{\left( {{x}^{2}}+9 \right)}\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

$\Rightarrow 5x=A\left( {{x}^{2}}+9 \right)+Bx+C\left( x+1 \right)$

$\Rightarrow 5x=A{{x}^{2}}+9A+B{{x}^{2}}+Bx+Cx+C$

On equating the coefficients of ${{x}^{2}},\,x$ and constant term, it can be obtained that,  

$A+B=0\,\,\,...\left( 2 \right)$ 

$B+C=5\,\,\,...\left( 3 \right)$

$9A+C=0\,\,\,...\left( 4 \right)$

And on solving these equations, the values of $A,\,B,\,C$ can be obtained as,

$A=\dfrac{-1}{2},\,B=\dfrac{1}{2},\,C=\dfrac{9}{2}$ respectively.

Now, from equation $\left( 1 \right)$ it can be clearly obtained that,

$\int{\dfrac{5x}{(x+1)\left( {{x}^{2}}+9 \right)}dx}=\int{\left\{ \dfrac{-1}{2\left( x+1 \right)}+\dfrac{x+9}{2\left( {{x}^{2}}+9 \right)} \right\}dx}$ 

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{2}\int{\dfrac{x}{{{x}^{2}}+9}dx}+\dfrac{9}{2}\int{\dfrac{1}{{{x}^{2}}+9}dx}$

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{4}\log \left| {{x}^{2}}+9 \right|+\left( \dfrac{9}{2} \right)\left( \dfrac{1}{3} \right){{\tan }^{-1}}\left( \dfrac{x}{3} \right)$

$=\dfrac{-1}{2}\log \left| x+1 \right|+\dfrac{1}{4}\log \left| {{x}^{2}}+9 \right|+\dfrac{3}{2}{{\tan }^{-1}}\left( \dfrac{x}{3} \right)+C$, where $C$ is any arbitrary constant.

Ans: The given expression is, $\dfrac{\sin x}{\sin \left( x-\alpha  \right)}$. 

 Now, substitute $x-\alpha =t$ 

$\therefore dx=dt$

$\therefore \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=\int{\dfrac{\sin \left( t+\alpha  \right)}{\sin t}dt}=\int{\dfrac{\sin t\cos \alpha +\cos tsin\alpha }{\sin t}dt}=\int{\cos \alpha +\cot tsin\alpha dt}$$\Rightarrow \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=t\cos \alpha +\sin \alpha \log \left| \sin t \right|+{{C}_{1}}=\left( x-\alpha  \right)\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|+{{C}_{1}}$ 

$\Rightarrow \int{\dfrac{\sin x}{\sin \left( x-\alpha  \right)}dx}=x\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|-\alpha \cos \alpha +{{C}_{1}}=x\cos \alpha +\sin \alpha \log \left| \sin \left( x-\alpha  \right) \right|+C$

 where ${{C}_{1}},\,C$ are any arbitrary constants and $C={{C}_{1}}-\alpha \cos \alpha $.

Ans: The given expression is, $\dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}$. 

 $\therefore \dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}=\dfrac{{{e}^{4\log x}}\left( {{e}^{\log x}}-1 \right)}{{{e}^{2\log x}}\left( {{e}^{\log x}}-1 \right)}={{e}^{2\log x}}={{x}^{2}}$

Now, integrate the given expression as shown below

$\therefore \int{\dfrac{{{e}^{5\log x}}-{{e}^{4\log x}}}{{{e}^{3\log x}}-{{e}^{2\log x}}}dx}=\int{{{x}^{2}}dx}=\dfrac{{{x}^{3}}}{3}+C$, where $C$ is any arbitrary constant. 

Ans: The given expression is, $\dfrac{\cos x}{\sqrt{4-{{\sin }^{2}}x}}$. 

 Now, substitute $\sin x=t$ 

$\therefore \cos xdx=dt$

$\therefore \int{\dfrac{\cos x}{\sqrt{4-{{\sin }^{2}}x}}dx}=\int{\dfrac{dt}{\sqrt{{{\left( 2 \right)}^{2}}-{{\left( t \right)}^{2}}}}}={{\sin }^{-1}}\left( \dfrac{t}{2} \right)+C={{\sin }^{-1}}\left( \dfrac{\sin x}{2} \right)+C$, where $C$ is any arbitrary constant. 

Ans: The given expression is, $\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}$. 

$\therefore \dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}=\dfrac{\left( {{\sin }^{4}}x-{{\cos }^{4}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{2}}x+{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x-{{\sin }^{2}}x{{\cos }^{2}}x}$

$=\dfrac{\left( {{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{2}}x-{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{2}}x\left( 1-{{\cos }^{2}}x \right)+{{\cos }^{2}}x\left( 1-{{\sin }^{2}}x \right)}$

$=\dfrac{-\left( -{{\sin }^{2}}x+{{\cos }^{2}}x \right)\left( {{\sin }^{4}}x+{{\cos }^{4}}x \right)}{{{\sin }^{4}}x+{{\cos }^{4}}x}=-\cos 2x$

Now, integrate the given expression as shown below

$\therefore \int{\dfrac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}dx}=\int{-\cos 2xdx}=\dfrac{-\sin 2x}{2}+C$, where $C$ is any arbitrary constant.