Class 12th

NCERT Exemplar problems are designed by the subject matter experts. It helps in deepening conceptual understanding and improving the application skills of the students. Preparing the NCERT exemplar Chapter 2 helps students in mastering complex topics like capacitors, electric potential, and energy storage which are important for the board exam preparations and entrance tests like JEE and NEET.

$\begin{array}{l}I=\int \frac{dx}{si{n}^{2}xco{s}^{2}x}\\ =\int \frac{1}{si{n}^{2}xco{s}^{2}x}dx\\ =\int \frac{si{n}^{2}x+co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx\\ =\int \frac{si{n}^{2}x}{si{n}^{2}xco{s}^{2}x}dx+\int \frac{co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx\\ =\int se{c}^{2}xdx+\int cose{c}^{2}x-dx]\\ =tanx-cotx+c\end{array}$

Therefore, the correct answer is B.

$\begin{array}{l}Put{x}^{10}+10^x=t\\ \Rightarrow \left(10x^9+10^{x}lo{g}^{e}10\right)dx=dt\\ I=\int \frac{10x^9+10^{x}lo{g}_{e}10}{x^{10}+10^x}dx\\ =\int \frac{dt}{t}=logt+C\\ =log\left(10^x+x^{10}\right)+c\end{array}$

Therefore, the correct answer is (D)

$\begin{array}{l}Put{x}^4=t\\ 4x^{3}dx=dt\\ I=\int \frac{x^{3}sin\left(ta{n}^{-1}{x}^4\right)}{1+x^8}dx\\ =\frac{1}{4}\int \frac{sin\left(ta{n}^{-1}t\right)}{1+t^2}\_\_\_\_\_(1)\\ Putta{n}^{-1}t=u\\ \Rightarrow \frac{1}{1+t^2}dt=du\end{array}$

From (1), we get

$\begin{array}{l}I=\int \frac{x^{3}sin\left(ta{n}^{-1}{x}^4\right)}{1+x^8}dx\\ =\frac{1}{4}{\displaystyle \int sin}udu\\ =\frac{1}{4}(-cosu)+C\\ =-\frac{1}{4}cos\left(ta{n}^{-1}t\right)+C\\ =-\frac{1}{4}cos\left(ta{n}^{-1}{x}^4\right)+C\end{array}$

$\begin{array}{l}I=\frac{(x+1)}{x}{(x+logx)}^2\\ =\left(1+\frac{1}{x}\right){(x+logx)}^2\\ Putx+logx=t\\ \Rightarrow 1+\frac{1}{x}dx=dt\\ I=\int \left(1+\frac{1}{x}\right){(x+logx)}^{2}dx\\ =\int t^{2}dt=\frac{t^3}{3}+C\\ =\frac{{(x+logx)}^3}{3}+C\end{array}$

$\begin{array}{l}Put\left(1+logx\right)=t\\ \Rightarrow \frac{1}{x}dx=dt\\ I=\int \frac{{(1+logx)}^2}{x}dx\\ =\int t^{2}dt\\ =\frac{t^3}{3}+C\frac{{(1+logx)}^3}{3}+C\end{array}$

Kindly go through the solution

$\begin{array}{l}I=\int \frac{1}{1-tanx}dx\\ =\int \frac{1}{1-\frac{sinx}{cosx}}dx=\int \frac{1}{\frac{cosx-sinx}{cosx}}dx\\ =\int \frac{cosx}{cosx-sinx}dx\\ =\frac{1}{2}\int \frac{2cosx}{cosx-sinx}dx\\ =\frac{1}{2}\int \frac{(cosx-sinx)+(cosx+sinx)}{(cosx-sinx)}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+\frac{1}{2}\int \frac{cosx+sinx}{cosx-sinx}dx\\ Putcosx-sinx=t\\ \Rightarrow \left(-sinx-cosx\right)dx=dt\\ I=\frac{x}{2}+\frac{1}{2}\int \frac{-dt}{t}\\ =\frac{x}{2}-\frac{1}{2}log|cosx-sinx|+C\end{array}$

$\begin{array}{l}\int \frac{1}{1+\frac{cosx}{sinx}}·dx=\int \frac{1}{\frac{sinx+cosx}{sinx}}dx\\ =\int \frac{sinx}{sin+cosx}·dx=\frac{1}{2}\int \frac{2sinx}{sinx+cosx}dx\\ =\frac{1}{2}\int (\frac{sinx+cosx)+(sinx-cosx)}{sinx+cosx}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+1\int \frac{\left(sinx-cosx\right)}{sinx+cosx}dx\\ =\frac{1}{2}{\displaystyle \int 1}·dx+1\int \frac{\left(sinx-cosx\right)}{sinx+cosx}dx\\ =\frac{1}{2}(x)+\frac{1}{2}\int \frac{sinx-cosx}{sinx+cosx}dx\\ Putsinx+cosx=t\\ \Rightarrow \left(cosx-sinx\right)dx=dt\\ =\frac{x}{2}+\frac{1}{2}\int \frac{-dt}{t}\\ =\frac{x}{2}-\frac{1}{2}log\left|t\right|+\; c\\ =\frac{x}{2}-\frac{1}{2}log|sinx+cosx|+\; c\end{array}$

$\begin{array}{l}Put1+cosx=t\\ -sinxdx=dt\\ I=\int \frac{sinx}{{(1+cosx)}^2}dx\\ =-\int \frac{dt}{t^2}=-\int t^{-2}dt\\ =\frac{1}{t}+C\\ =\frac{1}{1+cosx}+C\end{array}$