Exemplar Solution

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^{1/2}}{1+x^{3/4}}}dx\\ Putx=t^4\Rightarrow dx=4t^{3}dt\\ ={\displaystyle \int \frac{t^2.4t^3}{1+t^3}}dt=4{\displaystyle \int \frac{t^5}{1+t^3}}dt\\ =4{\displaystyle \int \left(t^2-\frac{t^2}{t^3+1}\right)}dt=4{\displaystyle \int t^2}dt-4{\displaystyle \int \frac{t^2}{t^3+1}}dt\\ LetI=I_1-I_2\\ Now{I}_1=4{\displaystyle \int t^2}dt=4.\frac{t^3}{3}+C_1=\frac{4}{3}{x}^{3/4}+C_1\\ and{I}_2=4{\displaystyle \int \frac{t^2}{t^3+1}}dt\\ Put{t}^3+1=z\Rightarrow 3t^{2}dt=dz\Rightarrow t^{2}dt=\frac{1}{3}dz\\ \therefore I_2=\frac{4}{3}{\displaystyle \int \frac{dz}{z}}=\frac{4}{3}log\left|z\right|+C_2=\frac{4}{3}log\left|t^3+1\right|+C_2\\ =\frac{4}{3}log\left|{\left(x\right)}^{3/4}+1\right|+C_2\\ \therefore I=I_1-I_2\\ =\frac{4}{3}{x}^{3/4}+C_1-\frac{4}{3}log\left|{\left(x\right)}^{3/4}+1\right|-C_2\\ \therefore I=\frac{4}{3}{x}^{3/4}-\frac{4}{3}log\left|{\left(x\right)}^{3/4}+1\right|+C_1-C_2\\ Hence,I=\frac{4}{3}\left[x^{3/4}-log\left|{\left(x\right)}^{3/4}+1\right|\right]+C.\left[?C_1-C_2=C\right]\end{array}$

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int ta{n}^{2}x}.se{c}^{4}xdx\\ ={\displaystyle \int ta{n}^{2}x.se{c}^{2}x.se{c}^{2}xdx=}{\displaystyle \int ta{n}^{2}x.\left(1+ta{n}^{2}x\right).se{c}^{2}xdx}\\ Puttanx=t,\therefore se{c}^{2}xdx=dt\\ \therefore I={\displaystyle \int t^2\left(1+t^2\right)dt}={\displaystyle \int \left(t^2+t^4\right)dt}={\displaystyle \int t^{2}dt+{\displaystyle \int t^4}}dt\\ =\frac{1}{3}{t}^3+\frac{1}{5}{t}^5=\frac{1}{3}ta{n}^{3}x+\frac{1}{5}ta{n}^{5}x+C\\ Hence,therequiredsolutionis\frac{1}{3}ta{n}^{3}x+\frac{1}{5}ta{n}^{5}x+C.\end{array}$