NCERT Solutions for Class 12 Maths Chapter 7 - Integrals - Exercise 7.6
Exercise 7.1 Solutions 22 Questions
Exercise 7.2 Solutions 39 Questions
Exercise 7.3 Solutions 24 Questions
Exercise 7.4 Solutions 25 Questions
Exercise 7.5 Solutions 23 Questions
Exercise 7.6 Solutions 24 Questions
Exercise 7.7 Solutions 11 Questions
Exercise 7.8 Solutions 6 Questions
Exercise 7.9 Solutions 22 Questions
Exercise 7.10 Solutions 10 Questions
Exercise 7.11 Solutions 21 Questions
Miscellaneous Exercise on Chapter 7 Solutions 44 Questions
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
$$\Rarr\space\text{I}= e^x\text{sec x}+\text{C}$$
