Integration By Parts (Part-7)| S N Dey| Class 12

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In the previous article, we have discussed the solutions of 10 Long Answer type questions of Integration By Parts Chapter of S N Dey mathematics Class 12. In this chapter, we will discuss few more problems of Long Answer Type Questions.

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

11(i)~~\displaystyle\int{e^x\frac{x^2+1}{(x+1)^2}}~dx

Solution.

\displaystyle\int{e^x\frac{x^2+1}{(x+1)^2}}~dx\\=\int{e^x\frac{x^2-1+2}{(x+1)^2}}~dx\\=\int{e^x \frac{x^2-1}{(x+1)^2}}~dx+2\int{e^x \frac{1}{(x+1)^2}}~dx\\=\int{e^x \frac{(x+1)(x-1)}{(x+1)^2}}~dx+2\int{e^x \frac{1}{(x+1)^2}}~dx\\=\int{e^x\frac{x-1}{x+1}}~dx+ 2\int{e^x \frac{1}{(x+1)^2}}~dx \\=I_1+2\int{e^x \frac{1}{(x+1)^2}}~dx \rightarrow(1)

I_1\\=\displaystyle\int{e^x\frac{x-1}{x+1}}~dx\\=\frac{x-1}{x+1} \int{e^x}~dx-\int{\left[\frac{d}{dx}\left(\frac{x-1}{x+1}\right) \int{e^x}~dx\right]}~dx \\= \frac{x-1}{x+1} \cdot e^x-\int{ \frac{(x+1) \cdot 1-(x-1) \cdot 1}{(x+1)^2} \cdot e^x}~dx\\=\frac{x-1}{x+1} \cdot e^x-\int{\frac{2}{(x+1)^2} \cdot e^x}~dx\\=\frac{x-1}{x+1} \cdot e^x-2\int{e^x \frac{1}{(x+1)^2}}~dx \rightarrow(2)

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

\displaystyle\int{e^x\frac{x^2+1}{(x+1)^2}}~dx\\=\frac{x-1}{x+1} \cdot e^x-2\int{e^x \frac{1}{(x+1)^2}}~dx+2\int{e^x \frac{1}{(x+1)^2}}~dx\\=\left(\frac{x-1}{x+1}\right)e^x+c.

11(ii)~~\displaystyle\int{e^x\left(\frac{x+2}{x+4}\right)^2}~dx

Solution.

\displaystyle\int{e^x\left(\frac{x+2}{x+4}\right)^2}~dx\\=\int{e^x\left(1-\frac{2}{x+4}\right)^2}~dx\\=\int{e^x \left[1-\frac{4}{x+4}+\frac{4}{(x+4)^2}\right]}~dx\\=\int{e^x}~dx-4\int{\frac{e^x}{x+4}}~dx+4\int{\frac{e^x}{(x+4)^2}}~dx\\=e^x-4I_1+4\int{\frac{e^x}{(x+4)^2}}~dx+c \rightarrow(1)

I_1\\=\displaystyle\int{\frac{e^x}{x+4}}~dx\\=\frac{1}{x+4} \int{e^x}~dx-\int{\left[\frac{d}{dx}\left(\frac{1}{x+4}\right) \int{e^x}~dx\right]}~dx\\=\frac{e^x}{x+4}+\int{\frac{e^x}{(x+4)^2}}~dx\rightarrow(2)

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

\displaystyle\int{e^x\left(\frac{x+2}{x+4}\right)^2}~dx\\=e^x-4\left[\frac{e^x}{x+4}+\int{\frac{e^x}{(x+4)^2}}~dx\right]+4\int{\frac{e^x}{(x+4)^2}}~dx+c\\=e^x-\frac{4e^x}{x+4}+c\\=\frac{xe^x+4e^x-4e^x}{x+4}+c\\=\frac{xe^x}{x+4}+c.

12.~\displaystyle\int{e^x\sin x}~dx

Solution.

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

I_1\\=\displaystyle\int{e^x\cos x}~dx\\=e^x \int{\cos x}~dx -\int{\left[\frac{d}{dx}(e^x) \int{\cos x}~dx\right]}~dx\\=e^x\sin x-\int{e^x\sin x}~dx\\=e^x\sin x-I+c_1 \rightarrow(2)

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

I=-e^x \cos x+e^x\sin x-I +c_1 \\ \therefore~I+I=e^x(-\cos x+\sin x) +c_1 \\ \text{or,}~ 2I=e^x(\sin x-\cos x) +c_1 \\ \text{or,}~ I=\frac 12 e^x(\sin x-\cos x)+c.~~[~c=c_1/2~]

13.~~ \displaystyle\int{e^{3x}\cos4x}~dx

Solution.

I\\=\displaystyle\int{e^{3x}\cos4x}~dx\\=e^{3x} \int{\cos4x}~dx-\int{\left[\frac{d}{dx}(e^{3x}) \int{\cos4x}~dx\right]}~dx\\=e^{3x}\left(\frac{\sin4x}{4}\right)-\int{3e^{3x}\cdot \frac{\sin4x}{4}}~dx\\=e^{3x}\left(\frac{\sin4x}{4}\right)-\frac 34\int{e^{3x}\sin4x}~dx\=e^{3x}\left(\frac{\sin4x}{4}\right)-\frac 34I_1\rightarrow(1)

I_1\\=\displaystyle\int{e^{3x}\sin4x}~dx\\=e^{3x}\int{\sin4x}~dx-\int{\left[\frac{d}{dx}(e^{3x}) \int{\sin4x}~dx\right]}~dx\\=e^{3x}\left(-\frac{\cos4x}{4}\right)-\int{3e^{3x} \cdot\left(-\frac{\cos4x}{4}\right) }~dx\\=e^{3x}\left(-\frac{\cos4x}{4}\right)+\frac 34\displaystyle\int{e^{3x}\cos4x}~dx \\=e^{3x}\left(-\frac{\cos4x}{4}\right)+\frac 34I+c_1\rightarrow(2)

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

I=e^{3x}\left(\frac{\sin4x}{4}\right)-\frac 34\left[e^{3x}\left(-\frac{\cos4x}{4}\right)+\frac 34I\right]+c_1\\~~ \\ \text{or,}~~ I=\frac 14e^{3x}\sin4x+\frac{3}{16}e^{3x}\cos4x-\frac{9}{16}I +c_1\\~~ \\ \therefore~ \left(1+\frac{9}{16}\right)I=\frac 14e^{3x}\sin4x+\frac{3}{16}e^{3x}\cos4x +c_1\\~~ \\ \text{or,}~~ \frac{25}{16}I=\frac 14e^{3x}\sin4x+\frac{3}{16}e^{3x}\cos4x+c_1\\~~ \\ \text{or,}~~ 25 I=4e^{3x}\sin4x+3e^{3x}\cos4x+16c_1\\~~ \\ \text{or,}~~ I=\frac{e^{3x}}{25}(4e^{3x}\sin4x+3e^{3x}\cos4x )+c.~~[~c=\frac{16}{25}c.~]

14.~~\displaystyle\int{2^x \sin x}~dx

Solution.

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

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

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

I=\frac{2^x\sin x}{\log 2}-\frac{1}{\log 2}\left[\frac{2^x \cos x}{\log 2}+\frac{1}{\log 2}I \right]\\~~ \\ \text{or,}~~ I+\frac{1}{(\log 2)^2}I=\frac{2^x \sin x}{\log 2}-\frac{2^x \cos x}{(\log 2)^2}\\~~ \\ \text{or,}~~ \left[1+\frac{1}{(\log 2)^2}\right]I=\frac{2^x}{(\log 2)^2}[\sin x \log 2-\cos x]+c_1\\~~ \\ \text{or,}~~ \left[\frac{(\log 2)^2+1}{(\log 2)^2}\right]I=\frac{2^x}{(\log 2)^2}[\sin x \log 2-\cos x]+c_1\\~~ \\ \therefore~ I=\frac{2^x}{(\log 2)^2+1}[\sin x \log 2-\cos x]+c.[*]

Note [*] : ~c=\frac{(\log 2)^2}{(\log 2)^2+1}c_1.

15.~~\displaystyle\int{3^{2x} \cos2x}~dx

Solution.

\displaystyle\int{3^{2x} \cos2x}~dx\\=\int{e^{\log 3^{2x}}\cos2x}~dx\\=\int{e^{2x \log 3}\cos2x}~dx\\=\int{e^{(2\log 3)x} \cdot \cos2x}~dx\\=\frac{e^{(2\log 3)x}}{(2\log 3)^2+2^2}[(2\log 3)\cos2x+2\sin2x]+c~[*]\\=\frac{e^{(\log 3^2)x}}{(\log 3^2)^2+2^2}[(\log 3^2)\cos2x+2\sin2x]+c\\=\frac{e^{(\log 9)x}}{(\log 9)^2+2^2}[(\log 9)\cos2x+2\sin2x]+c

Note[*]: ~\displaystyle\int{e^{ax}\cos bx}~dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+c

16.~~\displaystyle\int{e^x \cos^2x}~dx

Solution.

\displaystyle\int{e^x \cos^2x}~dx\\=\frac 12 \displaystyle\int{e^x (2\cos^2x)}~dx\\=\frac12\int{e^x(1+\cos2x)}~dx\\=\frac 12 \int{e^x}~dx+\frac 12\int{e^x \cos2x}~dx \\=\frac{e^x}{2}+\frac 12 \cdot \frac{\cos2x+2\sin2xe^x}{1^2+2^2}(1 \cdot \cos2x+2 \sin2x)~~[*]\\=\frac{e^x}{2}\left[1+\frac 15(\cos2x+2\sin2x)\right]+c.

Note[*]: ~\displaystyle\int{e^{ax}\cos bx}~dx=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)+c

17.~~ \displaystyle\int{3^x \cos^3x}~dx

Solution.

\displaystyle\int{3^x \cos^3x}~dx\\=\frac 14 \int{3^x(4\cos^3x)}~dx\\=\frac 14\int{3^x(\cos3x+3\cos x)}~dx\\=\frac 14\int{3^x\cos3x}~dx+\frac 34 \int{3^x\cos x}~dx\\=\frac 14I_1+\frac 34I_2 \rightarrow(1)

I_1\\=\displaystyle\int{3^x\cos3x}~dx\\=\int{e^{\log 3^x}\cos3x}~dx\\=\int{e^{x\log 3}\cos3x}~dx\\=\int{e^{(\log 3)x}\cos3x}~dx\\=\frac{e^{(\log 3)x}}{3^2+(\log 3)^2}[(\log 3)\cos3x+3\sin3x]\\=\frac{3^x}{9+(\log 3)^2}[(\log 3)\cos3x+3\sin3x]+c_1\rightarrow(2)

I_2\\=\displaystyle\int{3^x \cos x}~dx\\=\int{e^{\log 3^x}\cos x}~dx\\=\int{e^{x\log 3}\cos x}~dx\=\int{e^{(\log 3)x}\cos x}~dx\\=\frac{e^{(\log 3)x}}{1^2+(\log 3)^2}[(\log 3)\cos x+\sin x]\=\frac{e^{(\log 3)x}}{1+(\log 3)^2}[(\log 3)\cos x+\sin x]+c_2\rightarrow(3)

Hence, by ~(1),~(2),~(3)~ we get,

\displaystyle\int{3^x \cos^3x}~dx\\=\frac 14 \cdot\frac{3^x}{9+(\log 3)^2}[(\log 3)\cos3x+3\sin3x]\\~~+\frac 34 \cdot\frac{e^{(\log 3)x}}{1+(\log 3)^2}[(\log 3)\cos x+\sin x]+c.~~[~c=c_1+c_2~]

18.~~ \displaystyle\int{e^x\sin x \sin2x}~dx

Solution.

\displaystyle\int{e^x\sin x \sin2x}~dx\\=\frac 12 \int{e^x(2\sin x\sin2x)}~dx\\=\frac 12\int{e^x[\cos (2x-x)-\cos(2x+x) ]}~dx\\=\frac 12 \int{e^x(\cos x-\cos 3x)}~dx\\=\frac 12 \int{e^x \cos x}~dx-\frac 12 \int{e^x \cos 3x}~dx\\=\frac 12 \cdot \frac{e^x}{1^2+1^2}(1 \cdot \cos x+1 \cdot \sin x) -\frac 12 \cdot \frac{e^x}{1^2+3^2}(1 \cdot \cos3x+3 \cdot \sin3x)+c\\=\frac 14e^x(\cos x+\sin x)-\frac{1}{2} \cdot \frac{e^x}{10}(\cos3x+3\sin3x)+c.\\=\frac 14 e^x\left[\cos x+\sin x-\frac 15(\cos 3x+3\sin3x)\right]+c.

19(i)~~ \displaystyle\int{\frac{e^{\tan^{-1}x}}{(1+x^2)^2}}~dx

Solution.

I= \displaystyle\int{\frac{e^{\tan^{-1}x}}{(1+x^2)^2}}~dx\rightarrow(1)

~\text{let}~ \tan^{-1}x=z \\~~\\ \therefore~\frac{d}{dx}(\tan^{-1}x)=\frac{dz}{dz} \\~~\\ \text{or,}~ \frac{1}{1+x^2}~dx=dz\rightarrow(2)

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

I\\=\displaystyle\int{\frac{e^{\tan^{-1}x}}{1+x^2} \cdot \frac{1}{1+x^2}}~dx\\=\displaystyle\int{\frac{e^z}{1+\tan^2z}}~dz\\=\int{\frac{e^z}{\sec^2z}}~dz\\=\int{e^z\cos^2z}~dz\\=\frac 12 \int{e^z(2\cos^2z)}~dz\\=\frac 12 \int{e^z(1+\cos2z)}~dz\\=\frac 12 \int{e^z}~dz+\frac 12 \int{e^z \cos2z}~dz\\=\frac{e^z}{2}+\frac 12 \cdot \frac{e^z}{1^2+2^2}(\cos2z+2\sin2z)+c\\=\frac{e^z}{2}+\frac 12 \cdot \frac{e^z}{5}(\cos2z+2\sin2z)+c\\=\frac{e^z}{2}\left[1+\frac{\cos2z+2\sin2z}{5}\right]+c\\=\frac 12 e^{z}\left[1+\frac 15 \left(\frac{1-\tan^2z}{1+\tan^2z}+2 \times\frac{2\tan z}{1+\tan^2z}\right)\right]+c\\=\frac 12 e^{\tan^{-1}x}\left[1+\frac 15\left(\frac{1-x^2}{1+x^2}+\frac{4x}{1+x^2}\right)\right]+c\\=\frac 12 e^{\tan^{-1}x} \left[1+\frac{1-x^2+4x}{5(1+x^2)}\right]+c.

19(ii)~~ \displaystyle\int{\frac{e^{2\tan^{-1}x(1+x)^2}}{1+x^2}}~dx

Solution.

I= \displaystyle\int{\frac{e^{2\tan^{-1}x(1+x)^2}}{1+x^2}}~dx\rightarrow(1)

~\text{let}~ \tan^{-1}x=z \\~~\\ \therefore~\frac{d}{dx}(\tan^{-1}x)=\frac{dz}{dz} \\~~\\ \text{or,}~ \frac{1}{1+x^2}~dx=dz\rightarrow(2)

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

~I \\=\displaystyle\int{e^{2z}(1+\tan z)^2}~dx\\=\int{e^{2z}(1+2\tan z+\tan^2z)}~dz\\=\int{e^{2z}(1+\tan^2z+2\tan z)}~dz\\=\int{e^{2z}(\sec^2z+2\tan z)}~dz\\=\int{e^{2z}\sec^2z}~dz+2\int{e^{2z}\tan z}~dz\\=e^{2z}\int{\sec^2z}~dz-\int{\left[\frac{d}{dz}(e^{2z}) \int{\sec^2z}~dz\right]}~dz+2\int{e^{2z}\tan z}~dz\\=e^{2z}\tan z-2\int{e^{2z}\tan z}~dz+2\int{e^{2z}\tan z}~dz+c\\=e^{2z}\tan z+c\\=xe^{2\tan^{-1}x}+c

20.~~ \displaystyle\int{\csc^3x}~dx

Solution.

I\\=\displaystyle\int{\csc^3x}~dx\\=\int{\csc x(\csc^2x)}~dx\\=\csc x \int{\csc^2x}~dx-\int{\left[\frac{d}{dx}(\csc x) \int{\csc^2x}~dx\right]}~dx\\=\csc x(-\cot x)-\int{(-\csc x \cot x)(-\cot x)}~dx\\=-\csc x \cot x-\int{\csc x(\cot^2x)}~dx\\=-\csc x \cot x-\int{\csc x(\csc^2x-1)}~dx\\=-\csc x \cot x-\int{\csc^3x}~dx+\int{\csc x}~dx\\=-\csc x \cot x-I+\log|\tan(x/2)|+c_1\\~~ \\ \therefore~ 2I=-\csc x \cot x+\log|\tan(x/2)|+c_1 \\ \text{or,}~~ I=\frac 12[\log|\tan(x/2)|-\csc x\cot x]+c.~~[~c=c_1/2~]

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