# NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.6

Exercise 7.6

Direction (Q. 1 to 22) : Integrate the function.

1. x sin x

$$\textbf{Sol.}\space\text{Let}\space \text{I}=\int\text{sin x.x}\space dx$$

On taking x as first function and sin x as second function and integrating by parts, we get

$$\text{I} = x\int\text{sin x dx}-\\\int\bigg[\frac{d}{dx}(x)\int\text{sin x}\space dx\bigg]dx$$

= – x cos x + ∫ 1.cos x dx

$$\Rarr\space\text{I}=-\text{x cos x + sin x}+\text{C}$$

2. x sin 3x

Sol. Let I = ∫ x sin 3x dx

On taking x as first function and sin 3x as second function and integrating by parts, we get

$$\text{I = x}\int\text{sin 3x dx}-\\\int\bigg[\frac{d}{dx}(x)\int\text{sin 3x}\space dx\bigg]dx\\=\frac{-x cos 3x}{3}+\int\frac{\text{cos 3x}}{3}dx\\\bigg(\because\space \int \text{sin ax dx}=\frac{-\text{cos ax}}{a}\bigg)\\\Rarr\space \text{I}=\frac{-x cos 3x}{3}+\frac{1}{9}\text{sin 3x}+\text{C}$$

3. x2ex

Sol. Let I = ∫ x2exdx

On taking x2 as first function and ex as second function and integrating by parts, we get

$$\text{I}=x^2\int e^xdx-\int\bigg[\frac{d}{dx}(x^2)\int e^xdx\bigg]dx\\ x^2e^x-\int[ 2xe^x]dx$$

Again, integrating by parts, we get

$$\text{I}=x^2e^x-\\\begin{Bmatrix} 2x\int e^xdx-2\int\bigg[\frac{d}{dx}(x)\int e^x dx\bigg]dx\end{Bmatrix}$$

= x2ex – 2xex + 2∫exdx

= x2ex – 2xex + 2ex + C

$$\Rarr\space\text{I}=e^x(x^2-2x+2)+\text{C}$$

4. x log x

Sol. Let I = ∫ x log x dx. On taking log x as first function and x as second function and integrating by parts, we get

(∵ log function comes before algebraic function in ILATE)

$$\text{I} = \text{log}\space x\int x dx-\int\\\bigg[\frac{d}{dx}(log x)\int x dx\bigg]dx\\=\frac{x^2\text{log x}}{2}-\frac{1}{2}\int\frac{1}{x}x^2dx\\=\frac{x^2\text{log x}}{2}-\frac{1}{2}\int x dx\\\Rarr\space\text{I}=\frac{x^2 log x}{2}-\frac{1}{4}x^2+\text{C}$$

5. x log 2x

Sol. On taking log 2x as first function and x as second function and integrating by parts, we get

(∵ log function comes before algebraic function in ILATE)

I = ∫ x log 2x dx

$$=\text{log 2x}\int x dx-\\\int\bigg[\frac{d}{dx}(log 2x)\int x dx\bigg] dx\\=\frac{x^2}{2}\space\text{log 2x}-\\\int\bigg[\frac{1}{2x}×2×\frac{x^2}{2}dx\bigg]$$

$$=\frac{x^2\text{log 2x}}{2}-\frac{1}{2}\int x dx\\\Rarr\space \text{I}=\frac{x^2}{2}\text{log 2x}-\frac{1}{4}x^2+\text{C}$$

6. x2 log x

Sol. On taking log x as first function and x2 as second function and integrating by parts, we get

(∵ log function comes before algebraic function in ILATE)

I = ∫ x2 log x dx

$$= log \space x\int x^2 dx-\\\int\bigg[\frac{d}{dx}(log x)\int x^2 dx\bigg]dx\\=\frac{x^3}{3}\text{log x}-\int\bigg[\frac{1}{x}.\frac{x^3}{3}\bigg]dx\\=\frac{x^3}{3}\text{log x}-\frac{1}{3}\int x^2 dx\\=\frac{x^3}{3}\text{log x}-\frac{x^3}{9}+\text{C}$$

7. x sin–1 x

Sol. Let I = ∫ x sin–1 x dx

On taking sin–1 x as first function and x as second function and integrating by parts, we get

(∵ Inverse functions comes before, algebraic functions in ILATE)

$$\text{I}=\text{sin}^{\normalsize-1}\space x\int x dx-\\\int\bigg[\frac{d}{dx}(\text{sin}^{\normalsize-1}x)\int x dx\bigg]dx\\=\text{sin}^{\normalsize-1}x.\frac{x^2}{2}-\int\bigg[\frac{1}{\sqrt{1-x^2}}.\frac{x^2}{2}\bigg]dx\\=\frac{x^2}{2}.\text{sin}^{\normalsize-1}x+\int\bigg[\frac{1-1-x^2}{\sqrt{1-x^2}}.\frac{1}{2}\bigg]dx$$

(Add and subtract 1 in numerator of second terms)

$$=\frac{x^2}{2}.\text{sin}^{\normalsize-1}x-\frac{1}{2}\int\frac{1}{\sqrt{1-x^2}}dx+\\\frac{1}{2}\int\frac{1-x^2}{\sqrt{1-x^2}}dx\\=\frac{x^2}{2}.\text{sin}^{\normalsize-1}x-\frac{1}{2}\text{sin}^{\normalsize-1}x+\frac{1}{2}\int\sqrt{1-x^2}dx\\\Rarr\space \text{I}=\frac{x^2}{2}.\text{sin}^{\normalsize-1}x-\frac{1}{2}\text{sin}^{-1}x+\\\frac{1}{2}\bigg[\frac{1}{2}x\sqrt{1-x^2}+\frac{1}{2}\text{sin}^{\normalsize-1}x\bigg]+\text{C}\\\Rarr\space\text{I}=\text{sin}^{\normalsize-1}x\bigg[\frac{x^2}{2}-\frac{1}{4}\bigg]+\\\frac{1}{4}x\sqrt{1-x^2}+\text{C}$$

$$\Rarr\space\text{I}=\frac{\text{sin}^{\normalsize-1}x}{4}(2x^2-1)+\frac{1}{4}x\sqrt{1-x^2}+\text{C}$$

8. x tan–1 x

Sol. Let I = ∫ x tan–1 x dx

On taking tan–1 x as first function and x as second function and integrating by parts, we get

(∴ Inverse functions comes before algebraic functions in ILATE)

$$\therefore\space\text{I}=\text{tan}^{\normalsize-1}x\int x\space dx-\\\int\bigg[\frac{d}{dx}(tan^{\normalsize-1} x)\int x dx\bigg]dx\\=\text{tan}^{\normalsize-1}x.\frac{x^2}{2}-\int\bigg[\frac{1}{1+x^2}.\frac{x^2}{2}\bigg]dx\\=\frac{x^2}{2}.\text{tan}^{\normalsize-1}-\frac{1}{2}\int\bigg[\frac{x^2+1-1}{1+x^2}\bigg]dx$$

[Add and subtract 1 in numerator of second term]

$$=\frac{x^2}{2}.\text{tan}^{\normalsize-1}x-\\\frac{1}{2}\bigg[\int\frac{1+x^2}{1+x^2}dx-\int\frac{1}{1+x^2}dx\bigg]\\=\frac{x^2}{2}\text{tan}^{\normalsize-1}-\frac{1}{2}\bigg[\int 1 dx-\int\frac{1}{1+x^2}dx\bigg]$$

$$=\frac{x^2}{2}.\text{tan}^{\normalsize-1}x-\frac{1}{2}x+\frac{1}{2}\text{tan}^{\normalsize-1}x+\text{C}\\=\bigg(\frac{x^2+1}{2}\bigg)\text{tan}^{\normalsize-1}x-\frac{1}{2}x+\text{C}$$

9. x cos–1 x

Sol. Let I = ∫ x cos–1 x dx

Put cos–1 x = t

$$\Rarr\space x=\text{cos t}\Rarr\space dx=-\text{sin t dt}\\\therefore\space\text{I}=\int \text{x cos}^{\normalsize-1}xdx=\\-\int\text{ t cos t.sint dt}\\=-\frac{1}{2}\int t.2\text{sin t cos t dt}\\=-\frac{1}{2}\int \text{t. sin 2t dt}$$

(∵ 2 sin x cos x = sin 2x)

On taking t as first function and sin 2t as second function and integrating by parts, we get

$$\text{I}=-\frac{1}{2}\bigg[t \int\text{sin 2t dt}-\int\begin{Bmatrix}\frac{d}{dt}(t).\int\text{sin 2t dt}\end{Bmatrix}dt\bigg]$$

$$=-\frac{1}{2}\bigg[t\frac{(-\text{cos 2t})}{2} + \int 1.\frac{\text{cos 2t}}{2} dt\bigg]\\=\frac{1}{2}\text{t cos\space2t}-\frac{1}{4}\int\text{cos 2t dt}\\=\frac{1}{4}\text{t cos 2t}-\frac{1}{4}\frac{\text{sin 2t}}{2}+\text{C}\\=\frac{1}{4}\text{tcos 2t}-\frac{1}{8}\text{sin 2t}+\text{C}\\=\frac{1}{4}t(2\text{cos}^2t-1)-\frac{1}{8}.\text{2 sint cos t + C}$$

[∵ cos 2x = 2 cos2 x – 1 and sin 2x = 2 sin x cos x]

$$=\frac{1}{4}t(\text{2 cos}^2t-1)-\\\frac{1}{4}(1-\text{cos}^2t)^{1/2}\text{cos t}+\text{C}\\\begin{bmatrix}\because\space\text{sin}^2t+\text{cos}^2x=1\\\Rarr\text{sin x}=\sqrt{1-\text{cos}^2x}\end{bmatrix}$$

$$\therefore\space\int\text{x cos}^{\normalsize-1}x\space dx\\=\frac{1}{4}\text{cos}^{\normalsize-1}x(2x^2-1)-\\\frac{1}{4}(1-x^2)^{1/2}x+\text{C}\\\lbrack\text{put cos}^{\normalsize-1} x=t\space\text{and}\space\text{cos t=x}\rbrack\\=\frac{1}{4}(2x^2-1)\text{cos}^{\normalsize-1}x-\frac{1}{4}x\sqrt{1-x^2}+\text{C}$$

Note : Here, if we take cos–1 x as first function, then the integration becomes complicated. So, we convert inverse trigonometric function into trigonometric function to avoid the complication.

10. (sin–1 x)2

Sol. Let I = ∫ (sin–1 x)2dx

$$\text{Put sin}^{\normalsize-1}\theta\\\Rarr x=\text{sin}\space \theta\\\Rarr\space dx=\text{cos}\space\theta d\theta\\\therefore\space\text{I}=\int(\text{sin}^{\normalsize-1}x)^2dx=\int\theta^2\text{cos}\theta\space\text{d}\theta$$

On taking θ2 as first function and cos q as second function and integrating by parts, we get

$$=\theta^2\int\text{cos}\theta d\theta-\int\bigg[\frac{d}{d\theta}(\theta^2)\int\text{cos}\theta d\theta\bigg]d\theta$$

= θ2 sin θ – ∫ 2θ sin θ dθ

Again integrating by parts, we get

I = θ2 sinθ – 2[θ (– cos θ) – ∫ 1(– cos θ) dθ] + C

= θ2 sin θ + 2θ cos θ – 2∫ cos θ dθ + C

= θ2 sin θ + 2θ cos θ – 2 sin θ + C

$$=(\text{sin}^{\normalsize-1}x)^2x + 2\space\text{sin}^{\normalsize-1}x\\\sqrt{1-\text{sin}^2\theta}-2x+\text{C}\\\lbrack\text{Put}\space\theta=\text{sin}^{\normalsize-1}x\space\text{and sin}\space\theta=x\rbrack\\=x (\text{sin}^{\normalsize-1}x)^2+2\sqrt{1-x^2}\space\text{sin}^{\normalsize-1}x-2x+\text{C}$$

$$\textbf{11.}\space\frac{\textbf{ x\space{cos}}^{\normalsize-1}x}{\sqrt{\textbf{1-x}^\text{2}}}\\\textbf{Sol.}\space\text{Let I}=\int\frac{x \text{cos}^{\normalsize-1}x}{\sqrt{1-x^2}}dx\\\Rarr\space\text{I}=\int\text{cos}^{\normalsize-1}x.\frac{x}{\sqrt{1-x^2}}dx\\\text{Consider cos}^{\normalsize-1}x\space\text{as first function and}\\\frac{x}{\sqrt{1-x^2}}\space\text{as second function and}$$ integrating by parts, we get

$$\text{I}=\text{cos}^{\normalsize-1}x\int\frac{x}{\sqrt{1-x^2}}dx-\\\int\bigg[\frac{d}{dx}(\text{cos}^{\normalsize-1}x)\int\frac{x}{\sqrt{1-x^2}}dx\bigg]dx\\\text{Let}\space 1-x^2=t^2\\\Rarr\space -2x=2t\frac{d}{dx}\\\Rarr\space dx=-\frac{1 dt}{x}\\\therefore\space\text{I}=\text{cos}^{\normalsize-1}x\int\frac{x}{t}\bigg(\frac{-t}{x}\bigg)dt-\\\int\bigg[\frac{d}{dx}(\text{cos}^{\normalsize-1}x)\int\frac{x}{t}\bigg(\frac{-t}{x}\bigg)dt\bigg]dx$$

Let 1 – x2 = t2

$$\Rarr\space -2x=2t\frac{dt}{dx}\\\Rarr\space dx=-\frac{t\space dt}{x}\\\therefore\space\text{I = cos}^{\normalsize-1}x\int\frac{x}{t}\bigg(\frac{-t}{3}\bigg)dt-\\\int\bigg[\frac{d}{dx}(\text{cos}^{\normalsize-1}x)\int\frac{x}{t}\bigg(\frac{-t}{x}\bigg)dt\bigg]dx\\=\text{cos}^{\normalsize-1}x(-t)-\int\bigg[\frac{(-1)}{\sqrt{1-x^2}}(-t)dt\bigg]dx\\=-\sqrt{1-x^2}.\text{cos}^{\normalsize-1}x-\int\bigg[\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}\bigg]dx\\\begin{bmatrix}\because\space 1-x^2=t^2\\\Rarr\space t=\sqrt{1-x^2}\end{bmatrix}$$

$$=-\sqrt{1-x^2}.\text{cos}^{\normalsize-1}x-\int 1 dx\\=-\sqrt{1-x^2}.\text{cos}^{\normalsize-1}x-x+\text{C}$$

12. x sec2 x

Sol. Let I = ∫ x sec2 x dx

On taking x as first function and sec2 x as second function and integrating by parts, we get

$$\text{I}=x\int\text{sec}^2x dx-\\\int\bigg[\frac{d}{dx}(x)\int\text{sec}^2x dx\bigg]dx$$

= x tan x – ∫ tan x dx = x tan x – log|sec x| + C

$$\Rarr\space\text{I}= x\space tanx + \text{log}|\text{cos x}|+\text{C}\\\begin{bmatrix}\text{log}|\text{sec x}|=log \begin{vmatrix}\frac{1}{\text{cos x}}\end{vmatrix}= log 1-\\\text{log}|\text{cos x}|=-\text{log}|\text{cos x}|(\because\space\text{log =0})\end{bmatrix}$$

13. tan–1 x

Sol. Let I = ∫ 1 . tan–1 x dx

$$\Rarr\space\text{I}=\text{tan}^{\normalsize-1}x\int 1 dx-\\\int\bigg[\frac{d}{dx}(\text{tan}^{\normalsize-1}x)\int 1 dx\bigg]\\=\text{x tan}^{\normalsize-1}x-\int\frac{1}{1+x^2}.xdx\\\text{Let}\space 1+x^2=t\\\Rarr\space 2x=\frac{dt}{dx}\\\Rarr\space dx =\frac{dt}{2x}\\\therefore\space\text{I} = x\text{tan}^{-1}x-\int\frac{x}{t}.\frac{dt}{2x} \\= x\text{tan}^{\normalsize-1}x-\frac{1}{2}\int\frac{1}{t}dt$$

$$= x\text{tan}^{\normalsize-1}x-\frac{1}{2}\text{log}|t|+\text{C}\\= x\text{tan}^{\normalsize-1}x-\frac{1}{2}\text{log}|1+x^2|+\text{C}$$

14. x(log x)2

Sol. Let I = ∫ x(log x)2 dx

On taking (log x)2 as first function and x as second function and integrating by parts, we get

$$\text{I}=\text{(log x)}^2\int xdx-\\\int\bigg[\frac{d}{dx}(log \space x)^2\int x dx\bigg]dx\\=\text{(log x)}^2.\frac{x^2}{2}-\int\bigg[\frac{2 log x}{x}.\frac{x^2}{2}\bigg]dx\\=\frac{x^2}{2}(log x)^2-\int x log x dx$$

Again integrating by parts, we get

$$\text{I}=\frac{x^2}{2}(log x)^2-\begin{bmatrix}\text{log x}\int x dx-\\\int\begin{Bmatrix}\bigg(\frac{d}{dx}log x\bigg)\int x dx\end{Bmatrix}dx\end{bmatrix}+\text{C}$$

$$=\frac{x^2}{2}(log x)^2-\bigg[\frac{x^2}{2}\text{log x}-\int\frac{1}{x}.\frac{x^2}{2}dx\bigg]+\text{C}\\=\frac{x^2}{2}(log x)^2-\frac{x^2}{2}\text{log x}+\frac{1}{2}\int xdx+\text{C}$$

$$=\frac{x^2}{2}(log \space x)^2-\frac{x^2}{2}\text{log x}+\frac{x^2}{4}+\text{C}$$

15. (x2 + 1) log x

Sol. Let I = ∫ (x2 + 1) log x dx

On taking log x as first function and (x2 + 1) as second function and integrating by parts, we get

$$\text{I}=\text{log x}\int(x^2+1)dx-\\\int\bigg[\frac{d}{dx}(log x)\int(x^2+1)dx\bigg]dx\\\Rarr\space\text{I} = log \space x\bigg(\frac{x^3}{3} +x\bigg)-\\\int\frac{1}{x}.\bigg(\frac{x^3}{3}+x\bigg)dx\\\Rarr\space\text{I}=\bigg(\frac{x^3}{3}+x\bigg)\text{log x}-\int\bigg(\frac{x^2}{3}+1\bigg)dx\\=\bigg(\frac{x^3}{3}+x\bigg)\text{log x}-\frac{x^3}{9}-x+\text{C}$$

16. ex (sin x + cos x)

Sol. Let I = ∫ ex (sin x + cos x) dx

Let f(x) = sin x

$$\Rarr\space f'(x)=\text{cos x},\space\text{then,}\\\text{I}=\int e^x[f (x)+ f'(x)]dx$$

We know that ∫ ex (f(x) + f'(x)) dx = ex f(x)

∴ I = ex sin x + C

$$\textbf{17.\space}\frac{\textbf{xe}^\textbf{x}}{\textbf{(1+x)}^\textbf{2}}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{xe^x}{(1+x)^2}dx\\=\int e^x\frac{(x+1)-1}{(1+x^2)}dx\\=\int e^x\bigg[\frac{1}{(1+x)}-\frac{1}{(1+x)^2}\bigg]dx\\\text{Let}\space f(x)=\frac{1}{1+x}\\\Rarr\space f'(x)=-\frac{1}{(1+x)^2}$$

We know that

∫ ex(f(x) + f'(x)) dx = ex f(x)

$$\Rarr\space\text{I}=\int e^x\begin{Bmatrix}\frac{1}{1+x}-\frac{1}{(1+x)^2}\end{Bmatrix}dx\\=\frac{e^x}{1+x}+\text{C}$$

$$\textbf{18.}\space \textbf{e}^\textbf{x}\bigg(\frac{\textbf{1+sin x} }{\textbf{1 + cos x}}\bigg)\\\textbf{Sol.}\space\int e^x\bigg(\frac{1}{1+\text{cos\space x}}+\frac{\text{sin x}}{\text{1 + cos x}}\bigg)dx\\=\int e^x\begin{pmatrix}\frac{1}{\text{2 cos}^2\frac{x}{2}}+\frac{\text{2 sin}\frac{x}{2}\text{cos}\frac{x}{2}}{\text{2 cos}^2\frac{x}{2}}\end{pmatrix}dx\\\begin{bmatrix}\because\space \text{cos 2x = 2 cos}^2x-1\space\text{and}\\\text{sin 2x}=2\space\text{sin x cos x}\end{bmatrix}\\=\int e^x\begin{pmatrix}\frac{\text{sec}^2\frac{x}{2}}{2}+\text{tan}\frac{x}{2}\end{pmatrix}dx\\=\int e^x\bigg(\text{tan}\frac{x}{2}+\frac{1}{2}\text{sec}^2\frac{x}{2}\bigg)dx\\\text{Let}\space\text{f(x)}=\text{tan}\frac{x}{2}\\\Rarr\space f'(x)=\frac{\text{sec}^2\frac{x}{2}}{2}$$

$$\therefore\space\int e^x\bigg(\text{tan}\frac{x}{2}+\frac{\text{sec}^2\frac{x}{2}}{2}\bigg)dx\\= e^x\space\text{tan}\frac{x}{2}+\text{C}$$

[∵ ∫ ex {f(x) + f'(x)} dx = exf(x)]

$$\textbf{19.}\space \textbf{e}^\textbf{x}\bigg(\frac{\textbf{1}}{\textbf{x}}\textbf{-}\frac{\textbf{1}}{\textbf{x}^\textbf{2}}\bigg)\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int e^x\bigg(\frac{1}{x}-\frac{1}{x^2}\bigg)dx\\\text{Let}\space\text{f(x)}=\frac{1}{x}\\\Rarr\space f'(x)=-\frac{1}{x^2}$$

Here, given integral is of the form

∫ ex [f(x) + f'(x)] dx = ex f(x)

$$\therefore\space \text{I}=\frac{e^x}{x}+\text{C}$$

$$\textbf{20.}\space\frac{\textbf{x-3}}{\textbf{(x-1)}^\textbf{3}}\textbf{e}^\textbf{x}\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\frac{x-3}{(x-1)^3}e^x dx\\=\int\frac{x-1-2}{(x-1)^3}e^xdx\\=\int e^x\bigg(\frac{x-1}{(x-1)^3}-\frac{2}{(x-1)^3}\bigg)dx\\=\int e^x\bigg(\frac{1}{(x-1)^2}-\frac{2}{(x-1)^3}\bigg)dx\\\text{Let}\space f(x)=\frac{1}{(x-1)^2}\\\Rarr\space f'(x)=\frac{-2}{(x-1)^3}$$

$$\therefore\space\text{I}=\frac{e^x}{(x-1)^2}+\text{C}\\\lbrack\because\int e^x(f(x) + f'(x))dx=e^xf(x)\rbrack$$

21. e2x sin x

Sol. Let I = ∫ e2x sin x dx

On taking sin x as first function and e2x as second function and integrating by parts, we get

$$\int e^2x\text{sin x dx}=\\\text{sin x}.\int e^{2x} dx-\int\bigg[\frac{d}{dx}(sin x).\int e^{2x}dx\bigg]dx$$

$$=\text{sin}x.\frac{e^{2x}}{2}-\int\text{cos x}.\frac{e^{2x}}{2}dx\\=\text{sin x}.\frac{e^{2x}}{2}-\frac{1}{2}\int e^{2x}\text{cos x}dx\\\Rarr\space\int e^{2x}\space\text{sin x dx}=\text{sin x}.\frac{e^{2x}}{2}-\frac{1}{2}\text{I}_1\space\text{...(i)}$$

where, I1 = ∫ e2x cos x dx

On taking cos x as first function and e2x as second function and integrating by parts, we get

$$\text{I}_1=\text{cos x}.\int e^{2x}dx-\\\int\bigg[\frac{d}{dx}(cos \space x).\int e^{2x}dx\bigg]dx\\=\text{cos x}.\frac{e^{2x}}{2}-\int(-\text{sin x}).\frac{e^{2x}}{2}dx+\text{C}_1\\=\frac{e^{2x}cos x}{2}+\frac{1}{2}\int e^{2x}\space\text{sin x dx}+\text{C}_1$$

Putting value of I1 in equation (i), we get

$$\int\space e^{2x}\space\text{sin x}\space dx=\frac{e^{2x}sin x}{2}-\\\frac{1}{2}\bigg[\frac{e^{2x}cos x}{2}+\frac{1}{2}\int e^{2x}\space\text{sin x}dx + \text{C}_1\bigg]$$

$$=\frac{e^{2x}sin\space x}{2}-\frac{e^{2x}\text{cos x}}{4}-\\\frac{1}{4}\int e^{2x}\space\text{sin x dx}-\frac{1}{2}\text{C}_1\\\Rarr\space\int e^{2x}\space\text{sin x dx} + \frac{1}{4}\int e^{2x}\space\text{sin x dx}\\=\frac{e^{2x} sin x}{2}-\frac{e^{2x}cos x}{4}\\-\frac{1}{2}\text{C}_1$$

$$\Rarr\space \frac{5}{4}\int e^{2x}\space\text{sin x dx}=\\\frac{e^{2x} sin x}{2}-\frac{e^{2x} cos x}{4}-\frac{1}{2}\text{C}_1\\\Rarr\space\int e^{2x}\space\text{sin x dx}\\=\frac{4}{5}\bigg[\frac{e^{2x}sin x}{2}-\frac{e^{2x} cos x}{4}-\frac{1}{2}\text{C}_1\bigg]\\=\frac{2}{5}e^{2x}\text{sin x}-\frac{1}{5}e^{2x}\text{cos x}-\frac{2}{5}\text{C}_1\\=\frac{2}{5}e^{2x}\space\text{sin x}-\frac{1}{5}e^{2x}\text{cos x + C}\\\text{where, C}=-\frac{2}{5}\text{C}_1\\=\frac{e^{2x}}{5}(2 sin x - cos x)+\text{C}$$

$$\textbf{22.}\space\textbf{sin}^{\normalsize-1}\bigg(\frac{\textbf{2x}}{\textbf{1+x}^\textbf{2}}\bigg)\\\textbf{Sol.}\space\text{Let}\space\text{I}=\int\text{sin}^{\normalsize-1}\bigg(\frac{2x}{1+x^2}\bigg)dx$$

On putting x = tan θ, we get

$$\text{I}=\int\text{sin}^{\normalsize-1}\bigg(\frac{2 tan\theta}{1+tan^2\theta}\bigg)dx\\=\int\text{sin}^{\normalsize-1}(\text{sin}2 \theta)dx\\\bigg(\because\space \text{sin}2\theta=\frac{2 tan\theta}{1+\text{tan}^2\theta}\bigg)$$

= 2∫ θ dx = 2 ∫ tan–1 x dx

$$\lbrack\because\space x=tan\theta\Rarr\space =\text{tan}^{\normalsize-1}x\rbrack$$

$$= 2 \int \underset{\text{II}}{1}.\underset{\text{1}}{\text{tan}^{\normalsize-1}}\space x dx\\2\space\text{tan}^{\normalsize-1} x\int 1 dx-\\2\int\bigg[\frac{d}{dx}\text{tan}^{\normalsize-1}x\int 1 dx\bigg]dx\\\text{(using Integration by parts)}\\\Rarr\space \text{I}=2\text{tan}^{\normalsize-1}x.x-\\2\int\bigg[\frac{1}{1+x^2}.x\bigg]dx\\\text{Put 1+x}^2=t\\\Rarr\space 2x=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{2x}$$

$$\therefore\space \text{I = 2x tan}^{\normalsize-1}x-2\int\bigg[\frac{x}{t}.\frac{dt}{2x}\bigg]\\= 2 x\text{tan}^{\normalsize-1}x-\int\frac{1}{t}dt$$

= 2x tan–1 x – log|t| + C

$$\Rarr\space\text{I}=2x\space\text{tan}^{-1}x-\text{log}|1+x^2|+\text{C}$$

Choose the correct answer in the given questions.

$$\textbf{23.}\space\int \textbf{x}^\textbf{2}\textbf{e}^{\textbf{x}^\textbf{3}}\space \textbf{dx:}\\\textbf{(a)}\space\frac{\textbf{1}}{\textbf{3}}\textbf{e}^{\textbf{x}^\textbf{3}}\textbf{+ C}\\\textbf{(b)}\space\frac{\textbf{1}}{\textbf{3}}\textbf{e}^{\textbf{x}^{\textbf{2}}}\textbf{+ C}\\\textbf{(C)}\frac{\textbf{1}}{\textbf{2}}\textbf{e}^{\textbf{x}^{\textbf{3}}}\textbf{+ C}\\\textbf{(d)}\frac{\textbf{1}}{\textbf{2}}\textbf{e}^{\textbf{x}^{\textbf{2}}}\textbf{+ C}$$

$$\textbf{Sol.}\space \frac{1}{3}e^{x^{3}}+\text{C}\\\text{Let}\space \text{I}=\int x^2 e^{x^3}dx\\\text{Put}\space x^3=t\\\Rarr\space 3x^2=\frac{dt}{dx}\\\Rarr\space dx=\frac{dt}{3x^2}\\\therefore\space\text{I}=\frac{1}{3}\int e^tdt\\\Rarr\space\text{I}=\frac{1}{3} e^t+\text{C}\\=\frac{1}{3}e^{x^{3}}+\text{C}$$

$$\textbf{24.}\space\int \textbf{e}^\textbf{x}\textbf{sec x}(\textbf{1 + tan x})\textbf{dx}\space \textbf{equals}:\\\textbf{(a)\space }\textbf{e}^\textbf{X}\space\textbf{cos x + C}\\\textbf{(b)\space}\textbf{e}^\textbf{x}\space\textbf{sec x + C}\\\textbf{(C)\space}\textbf{e}^\textbf{x}\space\textbf{sin x + C}\\\textbf{(d)}\space \textbf{e}^\textbf{x}\space\textbf{tan x + C}\\\textbf{Sol.}\space (b)\space e^x\space\text{sec x + C}\\\text{Let}\space \text{I}=\int e^x\text{sec x}(1 + tan x)dx\\\Rarr\space \text{I =}\int e^x\text{sec x dx} + \int e^x\text{sec x tan x dx}\\\text{...(i)}\\\text{Now}\space\int e^x\space\text{sex x}dx$$

$$=\space\text{sec x}\int e^x dx-\int\bigg[\frac{d}{dx} sec x\int e^x dx\bigg]dx$$

= ex sec x – ∫ sec x tan x ex dx ...(ii)

On putting the value from equation (ii) in equation (i), we get

I = ex sec x – ∫ex sec x tan x dx + ∫ sec x tan x ex dx + C

$$\Rarr\space\text{I}= e^x\text{sec x}+\text{C}$$