Maths Integrals

Let $=\int \frac{cos2xdx}{{(sinx+cosx)}^2}$

$=\int \frac{co{s}^{2}x-si{n}^{2}x}{{(sinx+cosx)}^2}dx\{?cos2x=co{s}^{2}x-si{n}^{2}x$

$=\int \frac{(cosx+sinx)(cosx-sinx)}{{(sinx+cosx)}^2}dx\{?a^2-b^2=(x+b)(x-b)$

$=\int \frac{(cosx-sinx)}{sinx+cosx.}dx$

$=log|sinx+cosx|+c\{{\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=log|f(x)|+c}$

So, option B is correct.

Let $=\int \frac{dx}{e^x+e^{-x}}$

$=\int \frac{dx}{e^x+\frac{1}{e^x.}}=\int \frac{e^{x}dx}{e^x·e^x+1}$

$=\int \frac{e^{x}dx}{e^{2x}+1}$

Putting e^x = t

e^xdx = dt.

$\therefore =\int \frac{dt}{t^2+1.}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

Let $I={\displaystyle {\int }_0^1{e}^{2-3x}dx}$

We know that,

${\displaystyle {\int }_a^{b}f\left(x\right)}dx=\left(b-a\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(a\right)+f\left(a+h\right)+...+f\left(a\left(n-1\right)h\right)\right]$

Where, $h=\frac{b-a}{n}$

Here, $a=0,b=1$ and $f\left(x\right)e^{2-3x}$

$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$

$\therefore {\displaystyle {\int }_0^1{e}^{2-3x}dx}=\left(1-0\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(0\right)+f\left(0+h\right)+...+f\left(0+\left(n-1\right)h\right)\right]$

$\begin{array}{l}=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2+e^{2-3x}+...+e^{2-3\left(n-1\right)h}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{1+e^{-3h}+e^{-6h}+e^{-9h}+...+e^{-3\left(n-1\right)h}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-{\left(e^{-3h}\right)}^n}{1-\left(e^{-3h}\right)}\right\}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-e^{\frac{-3}{n}n}}{1-e^{\frac{-3}{n}}}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\end{array}$

$\begin{array}{l}=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\left(-\frac{1}{3}\right)\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]=\frac{e^2\left(e^{-3}-1\right)}{3}\underset{n\to \infty }{lim}\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]\\ =\frac{-e^2\left(e^{-3}-1\right)}{3}(1)\\ =\frac{-e^{-1}+e^2}{3}\\ =\frac{1}{3}\left(e^2-\frac{1}{e}\right)\end{array}$

Kindly go through the solution

Let I ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}·}2ta{n}^{3}xdx$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}ta{n}^2}xtanxdx$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\left(se{c}^{2}x-1\right)}tanxdx\left\{?se{c}^{2}x=ta{n}^{2}x+1\right\}$

$=2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}se{c}^2}xtanxdx-2{\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}tan}xdx$

$=\mathrm{2I}{}_1+2[log]{(cosx)}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

$=\mathrm{2I}{}_1+2\left[log\left(cos\frac{\pi }{4}\right)-log(cos0)\right]$

$=\mathrm{2I}{}_1+2(lo\mathrm{g1/\surd 2}$
$\frac{\mathrm{}}{}-log1)$

$=\mathrm{2I}{}_1+2(log·2^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-0)$

$I=\mathrm{2I}{}_1+2*\left(\frac{-1}{2}\right)log2=\mathrm{2I}{}_1-log2.\_\_\_\_(1).$

Where I${}_1={\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}se{c}^2}xtanxdx$

Let tan x = t =>sec²xdx = dt

When, x = 0, t = tan 0. = 0

$x=\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.,t=tan\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.=1$

${}_1={\displaystyle {\int }_0^{1}t}dt={\left[\frac{t^2}{2}\right]}_0^1=\frac{1}{2}-0=\frac{1}{2}$

So, Equation (1) becomes,

$=2*\frac{1}{2}-log2$

=1 - log 2.

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

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

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

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