Integration By Parts (Part-9)| S N Dey| Class 12 Examhoop

In this article, we will discuss the solutions of Long Answer Type Questions (31-42) in the chapter Integration By Parts (Ex-7) as given in the Chhaya Publication Book (Class 12) of aforementioned chapter of S N De book.

Integration By Parts , S N Dey Mathematics

Integration By Parts | S N Dey | Ex-7 | Long Answer Type Solution

$31.~~\displaystyle\int{\sqrt{3x^2+10x+3}}~dx$

Solution.

$I\\=\displaystyle\int{\sqrt{3x^2+10x+3}}~dx \\=\sqrt{3}\displaystyle\int{\sqrt{x^2+\frac{10}{3}x+1}}~dx \\=\sqrt{3}\displaystyle\int{\sqrt{x^2+2 \cdot x \cdot \frac 53+(5/3)^2+1-(5/3)^2}}~dx \\=\sqrt{3}\displaystyle\int{\sqrt{\left(x+\frac 53\right)^2+1-\frac{25}{9}}}~~dx \\=\sqrt{3}\displaystyle\int{\sqrt{\left(x+\frac 53\right)^2-\left(\frac 43\right)^2}}~dx \rightarrow(1)$

$\text{let}~~\left(x+\frac 53\right)=z \Rightarrow dx=dz\rightarrow(2)$

Hence, from $~(1),~(2)~$ we get,

$I\\=\sqrt{3}\displaystyle\int{\sqrt{z^2-\left(\frac 43\right)^2}}~dz\\=\sqrt{3}\left[\frac z2 \sqrt{z^2-\left(\frac 43\right)^2}-\frac{\left(\frac 43\right)^2}{2}\log \left|z+\sqrt{z^2-\left(\frac 43\right)^2}\right|\right] +c\\=\sqrt{3} \left[\frac{x+\frac 53}{2} \cdot \sqrt{\left(x+\frac 53\right)^2-\left(\frac 43\right)^2}-\frac{16}{9 \times 2}\log \left|\left(x+\frac 53\right)+\sqrt{x^2+\frac{10x}{3}+1}\right|\right]+c\\=\sqrt{3} \left[\frac 16(3x+5) \cdot \frac{\sqrt{3x^2+10x+3}}{\sqrt{3}}-\frac 89\log \left|\left(x+\frac 53\right)+\sqrt{x^2+\frac{10x}{3}+1}\right| \right]+c\\=\frac{3x+5}{6} \sqrt{3x^2+10x+3}-\frac{8\sqrt{3}}{9}\log \left|\left(x+\frac 53\right)+\sqrt{x^2+\frac{10x}{3}+1}\right|+c.$

$32.~~ \displaystyle\int{\frac{x^2+1}{\sqrt{1-x^2}}}~dx$

Solution.

$~ \displaystyle\int{\frac{x^2+1}{\sqrt{1-x^2}}}~dx\\=\int{\frac{2-(1-x^2)}{\sqrt{1-x^2}}}~dx\\=2\int{\frac{1}{\sqrt{1-x^2}}}~dx -\int{\frac{1-x^2}{\sqrt{1-x^2}}}~dx\\=2\sin^{-1}x-\left[\frac x2 \sqrt{1-x^2}+\frac 12 \sin^{-1}x\right]+c\\=\frac 32\sin^{-1}x-\frac x2\sqrt{1-x^2}+c\\=\frac 12(3\sin^{-1}x-x\sqrt{1-x^2})+c.$

$33.~~ \displaystyle\int{\frac{dx}{x-\sqrt{x^2-1}}}$

Solution.

$\displaystyle\int{\frac{dx}{x-\sqrt{x^2-1}}} \\=\int{\frac{(x+\sqrt{x^2-1})}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}}~dx\\=\int{\frac{(x+\sqrt{x^2-1})}{x^2-(x^2-1)}}~dx\\=\int{(x+\sqrt{x^2-1})}~dx\\=\int{x}~dx+\int{\sqrt{x^2-1}}~dx\\=\frac{x^2}{2}+\frac x2 \sqrt{x^2-1}-\frac 12 \log|(x+\sqrt{x^2-1})|+c.$

$34.~~ \displaystyle\int{(x+a)\sqrt{x^2-b^2}}~dx$

Solution.

$\displaystyle\int{(x+a)\sqrt{x^2-b^2}}~dx \\=\int{x\sqrt{x^2-b^2}}~dx+a\int{\sqrt{x^2-b^2}}~dx \\=I_1+aI_2 \rightarrow(1)$

$~I_1=\int{x\sqrt{x^2-b^2}}~dx \rightarrow(2)$

$\text{let}~~ x^2-b^2=z^2 \\ \therefore 2x~dx=2z~dz \\ \text{or,}~ x~dx=z~dz \rightarrow(3)$

Hence, by $~(2),~(3)~$ we get,

$I_1\\= \displaystyle\int{\sqrt{z^2} \cdot z}~dz\\=\int{z \cdot z}~dz\\=\int{z^2}~dz\\=\frac{z^3}{3}+c_1\\=\frac 13(x^2-b^2)^{\frac 32}+c_1 \rightarrow(4)$

$I_2\\=\displaystyle \int{\sqrt{x^2-b^2}}~dx\\=\frac{x\sqrt{x^2-b^2}}{2}-\frac{b^2}{2}\log|x+\sqrt{x^2-b^2}|+c_2 \rightarrow(5)$

Finally, by $~(1),~(4),~(5)~$ we get,

$\displaystyle\int{(x+a)\sqrt{x^2-b^2}}~dx\\=\frac 13(x^2-b^2)^{\frac 32}+c_1 +a\left(\frac{x\sqrt{x^2-b^2}}{2}-\frac{b^2}{2}\log|x+\sqrt{x^2-b^2}|+c_2\right)\\=\frac 13(x^2-b^2)^{\frac 32}+\frac 12 ax\sqrt{x^2-b^2}-\frac{ab^2}{2}\log|x+\sqrt{x^2-b^2}|+c~~[*]$

Note[*] : $~c=c_1+ac_2$

$35.~~ \displaystyle\int{(x-1)\sqrt{x^2+x+1}}~dx$

Solution.

$\displaystyle\int{(x-1)\sqrt{x^2+x+1}}~dx \\=\frac 12\int{(2x-2) \sqrt{x^2+x+1}}~dx\=\frac 12 \int{(2x+1-3)\sqrt{x^2+x+1}}~dx\=\frac 12 \int{(2x+1)\sqrt{x^2+x+1}}~dx-\frac 32\int{\sqrt{x^2+x+1}}~dx\=I_1-\frac 32I_2 \rightarrow(1)$

$~\text{let}~ x^2+x+1=z^2 \\ \therefore~(2x+1)~dx=2z~dz \rightarrow(2)$

By $~(1),~(2)~$ we get,

$I_1\\=\frac 12 \displaystyle \int{\sqrt{z^2} \cdot 2z}~dz\\=\int{z^2}~dz\=\frac 13 z^3+c_1\\=\frac 13(x^2+x+1)^{\frac 32}+c_1 \rightarrow(3)$

$~\sqrt{x^2+x+1}\\=\sqrt{x^2+2 \cdot x \cdot \frac 12+\left(\frac 12\right)^2+1-\left(\frac 12\right)^2}\\=\sqrt{\left(x+\frac 12\right)^2+1-\frac 14}\\=\sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \rightarrow(4)$

$I_2\\=\displaystyle\int{\sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}}~dx \\= \int{\sqrt{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}}~dz~\left[~z=x+\frac 12~\right] \\= \frac z2 \sqrt{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}+\frac{\left(\frac{\sqrt{3}}{2}\right)^2}{2} \log\left|z+\sqrt{z^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c_2 \\=\frac{x+\frac 12}{2} \sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} + \frac 38 \log \left|x+\frac 12+\sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\right|+c_2\rightarrow(5)$

Finally, by $~(1),~(4),~(5)~$ we get,

$\displaystyle\int{(x-1)\sqrt{x^2+x+1}}~dx\\= \frac 13(x^2+x+1)^{\frac 32}+c_1-\frac 32 \left[\frac{x+\frac 12}{2} \sqrt{\left(x+\frac 12\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} + \frac 38 \log \left|x+\frac 12+\sqrt{\left(x+\frac 12\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2}\right|+c_2\right]\\= \frac 13(x^2+x+1)^{\frac 32}-\frac{3(2x+1)}{8} \sqrt{x^2+x+1} -\frac{9}{16}\log \left|x+\frac 12+\sqrt{\left(x+\frac 12\right)^2+ \left(\frac{\sqrt{3}}{2}\right)^2}\right|+c~~[*]$

Note [*] : $~c=c_1-\frac{3}{2} c_2.$

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1	Chhaya Math Solution of Integration By Parts	Click here
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$36.~~ \displaystyle\int{xe^x \sin x}~dx$

Solution.

$I\\=\displaystyle\int{xe^x \sin x}~dx \\=x\int{e^x}~dx- \int{\left[\frac{d}{dx}(x) \int{e^x\sin x}~dx\right]}~dx\\=x \cdot \frac{e^x(\sin x-\cos x)}{2}-\int{1 \cdot \frac{e^x (\sin x-\cos x)}{2}}~dx\\=\frac{xe^x(\sin x-\cos x)}{2}-\frac 12 \int{e^x(\sin x-\cos x)}~dx \rightarrow(1)$

$\text{Now,}~~\displaystyle\int{e^x(\sin x-\cos x)}~dx\\=-\int{e^x(\cos x-\sin x)}~dx\\=-\int{e^x[f(x)+f'(x)]}~dx~~[*]\\=-e^xf(x)+c\\=-e^x \cos x+c \rightarrow(2)$

Note [*] : $~f(x)=\cos x \Rightarrow~~f'(x)=-\sin x$

Hence, by $~(1),~(2)~$ we get,

$~I=\frac{xe^x(\sin x-\cos x)}{2}+\frac 12e^x\cos x+c\\~~\\ ~\text{or,}~I=\frac{e^x}{2}[x(\sin x-\cos x)+\cos x]+c.$

$37.~~ \displaystyle\int{[\sin(\log x)+\cos(\log x)]}~dx$

Solution.

$\displaystyle\int{[\sin(\log x)+\cos(\log x)]}~dx \\=\int{\sin(\log x)}~dx+\int{\cos(\log x)}~dx\\=\sin(\log x)\int{dx}-\int{\left[\frac{d}{dx}(\sin(\log x)) \int{dx}\right]}~dx+\int{\cos(\log x)}~dx\\=x\sin(\log x)-\int{\cos(\log x) \cdot \frac 1x \cdot x~dx}+\int{\cos(\log x)}~dx\\= x\sin(\log x)-\int{\cos(\log x)}~dx+\int{\cos(\log x)}~dx+c\\=x\sin(\log x)+c.$

$38.~~ \displaystyle\int{(e^{\log x}+\sin x) \cos x}~dx$

Solution.

$\displaystyle\int{(e^{\log x}+\sin x) \cos x}~dx \\=\int{(x+\sin x) \cos x}~dx\\=\int{x\cos x}~dx+\int{\sin x\cos x}~dx\\=x\int{\cos x}~dx-\int{\left[\frac{d}{dx}(x) \int{\cos x}~dx\right]}~dx+\frac 12\int{\sin2x}~dx\\=x\sin x-\int{\sin x}~dx+\frac 12 \left(-\frac{\cos2x}{2}\right)+c\\=x\sin x+\cos x-\frac 14\cos 2x+c.$

$39.~~ \displaystyle\int{\sin(\log x)}~dx$

Solution.

$I\\=\displaystyle\int{\sin(\log x)}~dx\\=\sin(\log x)\int{dx}-\int{\left[\frac{d}{dx}(\sin(\log x)) \int{1 \cdot}~dx\right]}~dx\\=x\sin(\log x)-\int{\cos(\log x) \cdot \frac 1x \cdot x}~dx\\=x\sin(\log x)-\int{\cos(\log x)}\\=x\sin(\log x)-I_1 \rightarrow(1)$

$I_1\\= \displaystyle\int{\cos(\log x)}~dx\\=\cos(\log x)\int{dx}-\int{\left[\frac{d}{dx}(\cos(\log x)) \int{1 \cdot}~dx\right]}~dx\\=x\cos(\log x)+\int{\sin(\log x) \cdot \frac 1x \cdot x~dx}\\=x\cos(\log x)+\int{\sin(\log x)}~dx\\=x\cos(\log x)+I+c' \rightarrow(2)$

Hence, by $~(1),~(2)~$ we get,

$I=x\sin(\log x)-(x\cos(\log x)+I+c') \\ \therefore I+I=x\sin(\log x) -x\cos(\log x) -c' \\ \text{or,}~~ 2I=x[\sin(\log x)-\cos(\log x)] -c' \\ \text{or,}~~ I=\frac x2[\sin(\log x)-\cos(\log x)]+c. ~~[~c=-c'/2~]$

$41.~~$ Evaluate $~\displaystyle\int{e^{ax}\sin bx}~dx ~$ and $~\displaystyle\int{e^{ax}\cos bx}~dx~~$ by integrating each of them only once by parts.

Solution.

$42.~~$ Prove that, $~\displaystyle\int{\frac{e^x}{x^4}}~dx=-\frac{e^x}{3}\left[\frac{1}{x^3}+\frac{1}{2x^2}+\frac{1}{2x}\right]+\frac 16\int{\frac{e^x}{x}}~dx$

Solution.

$~\displaystyle\int{\frac{e^x}{x^4}}~dx\\=e^x \int{\frac{1}{x^4}}~dx-\int{\left[\frac{d}{dx}(e^x) \int{\frac{1}{x^4}}~dx\right]}~dx\\=e^x \cdot \frac{x^{-4+1}}{-4+1}-\int{e^x \cdot \left(\frac{x^{-4+1}}{-4+1}\right)}~dx\\=\frac{e^x}{-3x^3}+\frac 13 \int{\frac{e^x}{x^3}}~dx\\=\frac{e^x}{-3x^3}+\frac 13\left[e^x\int{\frac{1}{x^3}}~dx- \int{e^x \left(-\frac{1}{2x^2}\right)}~dx\right]\\=\frac{e^x}{-3x^3}+\frac{e^x}{3} \cdot \frac{x^{-3+1}}{-3+1}+\frac 13 \cdot \frac 12 \int{\frac{e^x}{x^2}}~dx\\=\frac{e^x}{-3x^3}-\frac{e^x}{6x^2}+\frac 16 \left[e^x\int{\frac{1}{x^2}}~dx- \int{\left[\frac{d}{dx}(e^x) \int{x^{-2}}~dx\right]}~dx\right]\\=\frac{e^x}{-3x^3}-\frac{e^x}{6x^2}+\frac{e^x}{6} \cdot \left(\frac{x^{-2+1}}{-2+1}\right)-\frac 16\int{e^x \cdot \left(\frac{x^{-2+1}}{-2+1}\right)}~dx\\=\frac{e^x}{-3x^3}-\frac{e^x}{6x^2}-\frac{e^x}{6x}+\frac 16 \int{\frac{e^x}{x}}~dx\\=-\frac{e^x}{3}\left[\frac{1}{x^3}+\frac{1}{2x^2}+\frac{1}{2x}\right]+\frac 16 \int{\frac{e^x}{x}}~dx$

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Integration By Parts | S N Dey | Ex-7 | Long Answer Type Solution

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