Class 12th

$Let,I={\displaystyle \int x}co{s}^{-1}xdx$

Putting cos^-1 x =θ=> x = cosθ=>dx = - sinθdθ.

$So,I={{\displaystyle \int }}^{\text{?}}cos\theta *\theta \cdot (-sin\theta )d\theta .$

$=-\frac{1}{2}{{\displaystyle \int }}^{\text{?}}\theta (2sin\theta cos\theta )d\theta$ {θsin 2θ = 2 sinθ cosθ}

$\begin{array}{l}=-\frac{1}{2}{{\displaystyle \int }}^{\text{?}}\theta sin2\theta d\theta .\\ =-\frac{1}{2}\left[\theta {{\displaystyle \int }}^{\text{?}}sin2\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d\theta }{d\theta }{{\displaystyle \int }}^{\text{?}}sin2\theta d\theta d\theta \right]\end{array}$

$\begin{array}{l}=-\frac{1}{2}\left[\frac{\theta (-cos2\theta )}{\theta }-{{\displaystyle \int }}^{\text{?}}\left(\frac{(-cos2\theta )}{2}\right)d\theta \right]\\ =\frac{\theta }{4}cos2\theta -\frac{1}{4}\cdot \frac{sin2\theta }{2}+C\\ =\frac{\theta }{4}cos2\theta -\frac{1}{8}(2sin\theta cos\theta )+C.\\ =\frac{\theta }{4}\left[2co{s}^2\theta -1\right]-\frac{1}{4}sin\theta cos\theta +C. \left\{?cos2\theta =2co{s}^2\theta -1\right\}.\\ \end{array}$$\begin{array}{l}\end{array}$

$\begin{array}{l}{\displaystyle \int x}ta{n}^{-1}xdx=ta{n}^{-1}x{\displaystyle \int x}dx-\int \frac{d}{dx}ta{n}^{-1}x·{\displaystyle \int xdx.dx}\\ =ta{n}^{-1}x·\frac{x^2}{2}-\int \frac{1}{1+x^2}·\frac{x^2}{2}dx.\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\int \frac{x^2}{1+x^2}dx\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\int \frac{\left(1+x^2\right)-1}{1+x^2}dx\\ =\frac{x^2}{2}·ta{n}^{-1}x-\frac{1}{2}\left[{\displaystyle \int \frac{\left(1+x^2\right)}{1+x^2}}dx-{\displaystyle \int \frac{dx}{1+x^2}}\right]\\ =\frac{x^2}{2}·ta{n}^{-1}x-\frac{1}{2}\left[{\displaystyle \int d}x-ta{n}^{-1}x\right]\\ =\frac{x^2}{2}ta{n}^{-1}x-\frac{1}{2}\left[x-ta{n}^{-1}x\right]+C.\\ =\frac{1}{2}\left[x^{2}ta{n}^{-1}x-x+ta{n}^{-1}x\right]+C.\\ =\frac{1}{2}\left[\left(x^2+1\right)ta{n}^{-1}x·-x\right]+C\end{array}$

Kindly go through the solution

$\begin{array}{l}\int x^2·logx·dx=logx·\int x^{2}dx-\int \frac{d}{dx}logx·\int x^{2}dxdx\\ =logx·\frac{x^3}{3}-\int \frac{1}{x}·\frac{x^3}{3}dx\\ =\frac{x^3}{3}·logx-\int \frac{x^2}{3}dx\\ =\frac{x^3}{3}·logx-\frac{1}{3}*\frac{x^3}{3}+c\\ =\frac{x^3}{3}logx-\frac{x^3}{9}+c.\end{array}$

$\begin{array}{l}{\displaystyle \int x}log2xdx=log2x·{\displaystyle \int x}dx-\int \frac{d}{dx}log2x{\displaystyle \int x}dxdx\\ =\frac{x^2}{2}·log2x-\int \frac{1}{2x}*\frac{d(2x)}{dx}*\frac{x^2}{2}dx+C\\ =\frac{x^2}{2}·log2x-\int \frac{1}{2x}*2*\frac{x^2}{2}{1}^{dx}+C\\ =\frac{x^2}{2}log2x-\frac{1}{2}{\displaystyle \int x}dx+c\\ =\frac{x^2}{2}log2x-\frac{1}{2}·\frac{x^2}{2}+c\\ =\frac{x^2}{2}log2x-\frac{x^2}{4}+c\end{array}$

$\begin{array}{l}{\displaystyle \int x}logxdx=logx{\displaystyle \int x}dx-\int \frac{d}{dx}logx·{\displaystyle \int x}dxdx\\ =logx*\frac{x^2}{2}-\int \frac{1}{x}*\frac{x^2}{2}dx\\ =\frac{x^2}{2}·logx-\frac{1}{2}{\displaystyle \int x}dx\\ =\frac{x^2}{2}logx-\frac{1}{2}*\frac{x^2}{2}+c\\ =\frac{x^2}{2}logx-\frac{x^2}{4}+c.\end{array}$

Answer – (b, c)

Explanation – According to the relation that p/q=r/s if galvanometer shows no deflection. In the above equation $\frac{R1}{R2}=\frac{R3}{R4}$ the value of R4 depends upon R1 and R2 . if the value of R1 and R2 will be feeble the value of R4 would be affected.

$\begin{array}{l}\int x^2{e}^{x}dx=x^2\int e^{x}dx-\int \frac{d{x}^2}{dx}\int e^{x}dxdx\\ =x^2·e^x-{\displaystyle \int 2}x{e}^{x}dx\\ =x^2{e}^x-2[x\int e^{x}dx-\int \frac{dx}{dx}\int e^{x}dxdx]\\ =x^2{e}^x-2\left[x{e}^x-\int e^{x}dx\right]\\ =x^2{e}^x-2x{e}^x+2e^x+c\\ =e^x\left[x^2-2x+2\right]+c\end{array}$

$\begin{array}{l}{\displaystyle \int x}sin3xdx=x{\displaystyle \int sin}3xdx-\int \frac{dx}{dx}{\displaystyle \int sin}3xdxdx.\\ =-x\frac{cos3x}{3}+\int \frac{cos3x}{3}\\ =\frac{-xcos3x}{3}+\frac{sin3x}{3*3}+c\\ =-\frac{x}{3}cos3x+\frac{1}{9}sin3x+c\end{array}$