Integrals

$=\int \frac{sinx}{sin(x-a)}dx$

Let x- a = t=> dx = dt.

$I\; =\int \frac{sin(t+a)}{sint}dt=\int \frac{sintcosa+costsina}{sint}dt$

$=cosa{\displaystyle \int d}t+sina{\displaystyle \int cot}tdt$

$=tcosa+sina·log\left|sint\right|+\; C$

$=(x-a)cosa+sinalog|sin(x-a)|+\; C$

$=xcosa+sinalog|sin(x-a)|+{}_{\mathrm{C1}}$

Where C₁ = C - a cos a

From (1), B = - A = $\left(-\frac{1}{2}\right)=\frac{1}{2}$

From (2) C = 5 - B = 5 $-\frac{1}{2}=\frac{10-1}{2}=\frac{9}{2}$

$\int \frac{5x}{(x+1)\left(x^2+9\right)}dx=\int \left\{\frac{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x+1}+\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.x+\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{x^2+9}\right\}dx$

$=-\frac{1}{2}\int \frac{dx}{x+1}+\frac{1}{2}\int \frac{x}{x^2+9}dx+\frac{9}{2}\int \frac{dx}{x^2+9}$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}\int \frac{9x}{x^2+9}dx+\frac{1}{2}-\int \frac{dx}{x^2+3^2}.$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}log\left(x^2+9\right)+\frac{9}{2}\times \frac{1}{3}ta{n}^{-1}\frac{x}{3}+\; C$

$=-\frac{1}{2}log|x+1|+\frac{1}{4}log\left(x^2+9\right)+\frac{3}{2}ta{n}^{-1}\frac{x}{3}+\; C$

Let I = $\frac{dx}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}={\displaystyle \int \frac{dx}{x^{\frac{1}{3}}\left(x^{\frac{1}{6}}+1\right)}}$

Putting $x^{\frac{1}{6}}=t\Rightarrow$ x = t⁶. dx = 6t⁵dt.

I = ${\displaystyle \int \frac{6t^{5}dt}{t^2\left(t+1\right)}dt}$

= $6{\displaystyle \int \frac{t^3}{\left(t+1\right)}dt.}$

= $6{\displaystyle \int \left[\frac{\left(t^2-t+1\right)\left(t+1\right)-1}{\left(t+1\right)}\right]dt}$

= $6\left\{{\displaystyle \int \left(t^2-t+1\right)}dt-{\displaystyle \int \frac{dt}{t+1}}\right\}$

= $6\left\{\frac{t^3}{3}-\frac{t^2}{2}+t-log|t+1|\right\}+\; C$

= $2x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-3t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}+6x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}-6log|x^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}+1|+\; C$

Let I = ${\displaystyle \int \frac{dx}{x^2{\left(x^4+1\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}$

= ${\displaystyle \int \frac{dx}{x^2{\left[x^4\left(1+\frac{1}{x^4}\right)\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}$ = ${\displaystyle \int \frac{dx}{x^2{x}^{4\times \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.{\left[1+\frac{1}{x^4}\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}}$

= ${\displaystyle \int \frac{dx}{x^{2+3}{\left[1+\frac{1}{x^4}\right]}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}$

= ${\displaystyle \int \frac{1}{x^5}{\left[1+\frac{1}{x^4}\right]}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}dx.}$

Putting $1+\frac{1}{x^4}=t\Rightarrow -4x^{-5}dx=dt$

$\frac{dx}{x^5}=\frac{dt}{-4}.$

I = ${\displaystyle \int t^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{dt}{-4}=-\frac{1}{4}\frac{t^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+1}}{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+1}+\; C}$

= $\frac{1}{4}\frac{t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+\; C$

= $t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+\; C$

= ${\left[1+\frac{1}{x^4}\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}+\; C$

Kindly go through the solution

Kindly go through the solution

$\frac{1}{2}.$

So, $\frac{1}{x-x^3}=\frac{1}{x}+\frac{1}{2}\frac{1}{\left(1-x\right)}-\frac{1}{2}\frac{1}{\left(1+x\right)}$

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

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

= $\frac{1}{2}\left[2log\left|x\right|-log|1-x|-log|1+x|\right]+\; C$

= $\frac{1}{2}\left[log{x}^2-log(1-x)(1+x)\right]+\; C$

= $\frac{1}{2}log\left(\frac{x^2}{1-x^2}\right)+\; C$

Let I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{4+3sinx}{4+3cosx}dx.$ ______(i)

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{4+3sin\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)}{4+3cos\left(\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.-x\right)}dx.$

I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\frac{4+3cosx}{4+3sinx}dx$ ______(ii)

Adding (i) & (ii) we get,

2I = ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left[log\left(\frac{4+3sinx}{4+3cosx}\right)+log\left(\frac{4+3cosx}{4+3sinx}\right)\right]}dx$

= ${\displaystyle {\int }_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}log}\left(\frac{4+3sinx}{4+3cosx}\times \frac{4+3sinx}{4+3cosinx}\right)dx$

= ${{\displaystyle \int }}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ log 1 dx

= ${{\displaystyle \int }}^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ × 0 dx

= 0

I = 0.

?Option (C) is correct

Kindly go through the solution

Given, f(x) = f(a – x)

g(x) + g(a – x) = 4.

Let I = ${\displaystyle {\int }_0^{a}f}(x)g(x)dx.$ ____(1)

= ${{\displaystyle \int }}^a$ f(x) g (a – x)dx

I = ${{\displaystyle \int }}^a$ f(x) g(a – x)dx______(2)

Adding (1) and (2),

2I = ${{\displaystyle \int }}^a$ [f(x) g(x) + f(x) g(a + x)]dx

= ${{\displaystyle \int }}^a$ f(x) [g(x) + g(a + x)]dx

2I = ${{\displaystyle \int }}^a$ f(x) 4 dx

I = $\frac{4}{2}{{\displaystyle \int }}^a$ f(x)dx = 2 ${{\displaystyle \int }}^a$ f