Integrals

LHS = I$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}si{n}^3}xdx.\{sin\mathrm{3A}=3si\mathrm{nA}-4si{n}^{\mathrm{3\; A}}$

$={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{4}}(3sinx-sin3x)dx.si{n}^{\mathrm{3A}}=\frac{1}{4}(3si\mathrm{nA}-sin\mathrm{3A})$

$=\frac{1}{4}\left[3{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}sin}xdx-{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$q$}\right.}sin}3xdx\right]$

$=\frac{1}{4}\left\{3{[-cosx]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left[\frac{-cos3x}{3}\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right\}$

$=\frac{3}{4}(-cos\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.+cos0)-\frac{1}{12}\left(-cos\frac{3\pi }{2}+cos3*0\right)$

$=\frac{3}{4}(-0+1)-\frac{1}{12}(-0+1)$

$=\frac{3}{4}-\frac{1}{12}=\frac{9-1}{12}=\frac{8}{12}=\frac{2}{3}$

$={\displaystyle {\int }_{-1}^1{x}^{17}}·co{s}^{4}xdx$

Here f (x) = x¹⁷ cos4x

f ( -x) = ( -x)¹⁷ cos⁴ ( -x)

= -x¹⁷ cos⁴x

= f (x)

i e, odd fxⁿ

As ${\displaystyle {\int }_a^{a}f}(x)dx=0$ for odd fxⁿ

therefore, I = 0.

LHS= ${\displaystyle {\int }_0^{1}x}{e}^{x}dx=x{\displaystyle {\int }_0^1{e}^x}dx-{\displaystyle {\int }_0^1\frac{dx}{dx}}\int e^{x}dxdx$

$={\left[x{e}^x\right]}_0^1-{\displaystyle {\int }_0^1{e}^x}dx$

$=\left[1e^1-0*e^0\right]-{\left[e^x\right]}_0^1$

$=\left(e^1-0\right)-\left(e^1-e^0\right)$

= e-e + e⁰

= e⁰ = 1=RHS

Let $I={\displaystyle {\int }_1^3\frac{dx}{x^2(x+1)}}.$

The integrand is of the form.

1 = Ax (x + 1) + B (x + 1) + Cx²

= A (x² + x) + B (x + 1) + Cx²

Comparing the coefficients,

A + C = 0 ____ (1)

A + B = 0 ______ (2)

B = 1 ________ (3)

Putting Equation (3) in (2),

A + 1 = 0

A = -1.

and putting value of A in Equation (1),

-1 + C = 0

C = 1

$\frac{1}{x^2(x+1)}=\frac{-1}{x}+\frac{1}{x^2}+\frac{1}{x+1}$

$\therefore I={\displaystyle {\int }_1^3\frac{dx}{x^2(x+1)}}={\displaystyle {\int }_1^3\frac{-dx}{x}}+{\displaystyle {\int }_1^3\frac{dx}{x^2}}+{\displaystyle {\int }_1^3\frac{dx}{x+1}}$

$=-{[log|x|]}_1^3+{\left[\frac{x^{-2+1}}{-2+1}\right]}_1^3+{[log?x+1]}_1^3$

$=[-log3+log1]-{\left[\frac{1}{x}\right]}_1^3+\left[log|3+1|-log|1+1|\right]$

$=-log3+0-\left[\frac{1}{3}-1\right]+log4-log2$

$=-log3-\frac{(1-3)}{3}+log{2}^2-log2$

$=-log3-\frac{(-2)}{3}+2log2-log2$

$=log2-log3+\frac{2}{3}$

$=\frac{2}{3}+log\frac{2}{3}$

Hence proved.

Let I = ${\displaystyle {\int }_1^4[|x-1|+|x+2|+|x-3|]}dx$

I = ${\displaystyle {\int }_1^4|}x-1|dx+{\displaystyle {\int }_1^4|}x+2|dx+{\displaystyle {\int }_1^4|}x-3|dx$

I = I₁ + I₂ + I₃______(1)

So, I₁ = ${{\displaystyle \int }}^4$ |x – 1| dx . $\{\begin{array}{l}(x-1)x-1>0\Rightarrow x>1\\ (x-1)x-1<0\Rightarrow x<1\end{array}$

= ${\displaystyle {\int }_1^4(}x-1)dx$

= ${\left[\frac{x^2}{2}-x\right]}_1^4=\left(\frac{4^2}{2}-\frac{1^2}{2}\right)-(4-1)$

= $\frac{15}{2}-3=\frac{15-6}{2}=\frac{9}{2}.$

I₂ = ${\displaystyle {\int }_1^4|}x-2|dx$ $\{\begin{array}{l}x-2x-2>0,x>2\\ -(x-2)x-2<0,x<2.\end{array}$

= ${\displaystyle {\int }_1^2-}(x-2)dx+{\displaystyle {\int }_2^4(}x-2)dx$

= $-{\left[\frac{x^2}{2}-2x\right]}_1^2+{\left[\frac{x^2}{2}-2x\right]}_2^4$

= $-\left[\left(\frac{2^2}{2}-\frac{1}{2}\right)-(2*2-2*1)\right]+\left[\left(\frac{4^2}{2}-\frac{2^2}{2}\right)-(2*4-2*2)\right]$

= $-\left[\frac{4-1}{2}-(4-2)\right]+[(8-2)-(8-4)]$

= $-\frac{3}{2}+2+6-4$

= $\frac{-3+4+12-8}{2}=\frac{5}{2}$

 I₃ = ${\displaystyle {\int }_1^4|}x-3|dx$ $\{\begin{array}{l}(x-3)x-3>0,x>3\\ -(x-3)x-3<0,x<3\end{array}$

= ${\displaystyle {\int }_1^3-}(x-3)dx+{\displaystyle {\int }_3^4(}x-3)dx$

= $-{\left[\frac{x^2}{2}-3x\right]}_1^3+{\left[\frac{x^2}{2}-3x\right]}_3^4$

= $-\left[\left(\frac{3^2}{2}-\frac{1^2}{2}\right)-(3·3-3·1)\right]+\left[\left(\frac{4^2}{2}-\frac{3^2}{2}\right)-(3*4-3*3)\right]$

= $-\left[\frac{8}{2}-6\right]+\left[\frac{7}{2}-3\right]$

= $-4+6+\frac{7}{2}-3=\frac{-8+12+7-6}{2}=\frac{5}{2}$

Hence Equation (1) becomes

I = $\frac{9}{2}+\frac{5}{2}+\frac{5}{2}$

I = $\frac{19}{2}$

$\begin{array}{l}2I={\displaystyle {\int }_0^n\frac{\pi tanx}{secx+tanx}dx\Rightarrow 2I=\pi {\displaystyle {\int }_0^n\frac{\frac{sinx}{cosx}}{\frac{1}{cosx}+\frac{sinx}{cosx}}}}dx\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }\frac{sinx+1-1}{1+sinx}dx}\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }1.dx-\pi {\displaystyle {\int }_0^{\pi }\frac{1}{1+sinx}}dx}\\ \Rightarrow 2I=\pi {\displaystyle {\int }_0^{\pi }1.dx-\pi {\displaystyle {\int }_0^{\pi }\frac{\left(1-sinx\right)}{\left(1+sinx\right)\left(1-sinx\right)}}dx}\end{array}$

$\begin{array}{l}\Rightarrow 2I=\pi {\left[x\right]}_0^{\pi }-\pi {\displaystyle {\int }_0^{\pi }\frac{1-sinx}{co{s}^{2}x}dx}\\ \Rightarrow 2I={\pi }^2-\pi {\displaystyle {\int }_0^{\pi }\left(se{c}^{2}x-tanxsecx\right)}dx\\ \Rightarrow 2I={\pi }^2-\pi {\left[tanx-secx\right]}_0^{\pi }\end{array}$

$\begin{array}{l}\Rightarrow 2I={\pi }^2-\pi \left[tan\pi -sec\pi -tan0+sec0\right]\\ \Rightarrow 2I={\pi }^2-\pi \left[0-\left(-1\right)-0+1\right]\\ \Rightarrow 2I={\pi }^2-2\pi \\ \Rightarrow 2I=\pi \left(\pi -2\right)\\ I=\frac{\pi }{2}\left(\pi -2\right)\end{array}$

Let I = ${\displaystyle {\int }_0^{\pi /2}sin}2xta{n}^{-1}(sinx)dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}2}sinxcosxta{n}^{-1}(sinx)dx$

Putting sin x = t =>cos xdx = dt.

whenx = 0, t = sin 0 = 0.

$x=\pi /2t=sin\pi /2=1$

? I = ${\displaystyle {\int }_0^{1}2}·tta{n}^{-1}(t)dt$

= $2\left[ta{n}^{-}t{\displaystyle {\int }_0^{1}t}dt-{\displaystyle {\int }_0^1\frac{d}{dt}}ta{n}^{-t}{\displaystyle \int t}dtdt\right]$

= $2\left\{{\left[ta{n}^{-1}t*\frac{t^2}{2}\right]}_0^1-{\displaystyle {\int }_0^1\frac{1}{1+t^2}}*\frac{t^2}{2}dt\right\}$

= $2\left\{\left[ta{n}^{-1}(1)*\frac{1}{2}-ta{n}^{-1}(0)*\frac{0}{2}\right]-\frac{1}{2}{\displaystyle {\int }_0^1\frac{\left(1+t^2\right)-1}{1+t^2}}dt\right\}$

= $2\left[\frac{\pi }{8}-0\right]-\frac{2}{2}\left\{{\displaystyle {\int }_0^1\frac{1+t^2}{1+t^2}}dt-{\displaystyle {\int }_0^1\frac{dt}{1+t^2}}\right\}$

= $\frac{\pi }{4}-{\displaystyle {\int }_0^{1}d}t+{\left[ta{n}^{-1}t\right]}_0^1$

= $-\frac{\pi }{4}-{[t]}_0^1+\left[ta{n}^{-1}(1)-ta{n}^{-1}(0)\right]$

= $\frac{\pi }{4}-1+\frac{\pi }{4}=2*\frac{\pi }{4}-1=\frac{\pi }{2}-1$

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{sinx+cosx}{9+16si{n}^{2}x}}dx$

Let sin x – cos x = t. =>(cosx + sin x) dx = dt.

and (sin x – cos x)² = t²

sin²x + cos²x – 2 sin x cos x = t²

1 – sin2x = t².

sin2t = 1 - t².

When x = 0, t = sin 0 – cos 0 = –1$$

? I = ${\displaystyle {\int }_{-1}^0\frac{dt}{9+16\left(1-t^2\right)}}={\displaystyle {\int }_{-1}^0\frac{dt}{9+16-16t^2}}$

= ${\displaystyle {\int }_{-1}^0\frac{dt}{25-16t^2}}$

= ${\displaystyle {\int }_{-1}^0}\frac{dt}{16\left(\frac{25}{16}-t^2\right)}$

= $\frac{1}{16}{\displaystyle {\int }_{-1}^0\frac{dt}{{\left(\frac{5}{4}\right)}^2-t^2}}$

= $\frac{1}{16}{\left[\frac{1}{2*\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)}log\left|\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+t}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.-t}\right|\right]}_{-1}^0\left\{{\displaystyle \int \frac{dx}{a^2-x^2}=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|}\right\}$

= $\frac{1}{16}*\frac{4}{2*5}{\left[log\left|\frac{5+4t}{5-4t}\right|\right]}_{-1}^0$

= $\frac{1}{40}\left[log\frac{5+4*0}{5-4*0}-log\frac{5+4(-1)}{5-4(-1)}\right]$

= $\frac{1}{40}\left[log\frac{5}{5}-log\frac{1}{9}\right]$

= $\frac{-1}{40}\left[log1-log{9}^{(-1)}\right]$

= $\frac{1}{40}[0-(-1)log9]$

= $\frac{1}{40}log9$

= $\frac{1}{40}log{3}^2=\frac{2}{40}log3$

= $\frac{1}{20}log3$

Kindly go through the solution

Kindly go through the solution