Class 12th

Preparing from the NCERT Exemplar Chapter 2 solutions will help in providing a strong concept foundation as it covers a wide range of question types from basic to advanced. If one practices the exemplar, they can score high in Board exams and entrance exams. However, it is advisable to first start from the NCERT textbook for the theory of the concepts and later on practice exemplars to boost the understanding and solve various types of questions based on these concepts. The exemplars are aligned with the CBSE syllabus and offer great study material for strengthening concept understanding and improving problem-solving accuracy.

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

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

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