Class 12th

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

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$

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

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

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

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

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*\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

Yes, PW Institute of Innovation, Noida accepts Class 12 marks for admission. The college offers BTech, BBA, and BVoc programmes at the UG levels. Candidates looking for course admission must complete the basic educational qualification, i.e., Class 12. The application process at the PW Institute of Innovation, Noida, is conducted online. 

Yes, PW Institute of Innovation, Pune accepts Class 12 marks for admission. The college offers BTech, BBA, and BVoc programmes at the UG levels. Candidates looking for course admission must complete the basic educational qualification, i.e., Class 12. The application process at the PW Institute of Innovation, Pune, is conducted online. 

$\frac{1}{2}.$

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}*\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