Class 12th

This is a Long Answer Type Questions as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2gh}$

$=\sqrt[]{2*10*500}=100m/s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x = $\sqrt[]{u^2-u_y^2}=\sqrt[]{{125}^2-{100}^2}=75m/s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2*500}{10}}=10s$

Time taken to reach the ground in 10 sec = 75 (10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

$Let,I={\displaystyle \int \frac{x{e}^x}{{(x+1)}^2}}dx={\displaystyle \int \frac{x+1-1}{{(x+1)}^2}}{e}^{x}dx$

$={\displaystyle \int e^x}\left[\frac{(x+1)}{{(x+1)}^2}+\frac{(-1)}{{(x+1)}^2}\right]dx.$ is of the form

e^x [f(x) + f(x)] dx

$\begin{array}{l}Where,f(x)=\frac{x+1}{{(x+1)}^2}=\frac{1}{x+1}\\ So,f^{\prime }(x)=\frac{d{(x+1)}^{-1}}{dx}=\frac{(-1)}{{(x+1)}^2}.\\ \therefore {\displaystyle \int \frac{x^{\cdot }\cdot e^x}{\left(1+x^2\right)}}dx=e^x\left[\frac{1}{x+1}\right]+\; C\end{array}$

∴f (x) = sin x

f (x) = cos x.

e^x [f (x) + f (x)] dx = e^xf (x) + C

${\displaystyle \int e^x(sinx+cosx)dx=e^x}sinx+c$

$\begin{array}{l}{\displaystyle \int \left(x^2+1\right)}logxdx=logx{\displaystyle \int \left(x^2+1\right)dx-{\displaystyle \int \left(x^2+1\right)dx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int \left(x^2+1\right)dxdx}}}}\\ =logx\cdot \left[\frac{x^3}{3}+x\right]-{\displaystyle \int \frac{1}{x}*\left[\frac{x^3}{3}+x\right]dx}\\ =\left[\frac{x^3}{3}+x\right]logx-{\displaystyle \int \left[\frac{x^2}{3}+1\right]dx}\\ =\left[\frac{x^3}{3}+x\right]logx-\frac{x^3}{3*3}-x+\; C\\ =\left[\frac{x^3}{3}+x\right]logx-\frac{x^3}{9}-x+\; C\end{array}$

$\begin{array}{l}{\displaystyle \int x{\left(logx\right)}^2=\cdot {\left(logx\right)}^2{\displaystyle \int xdx}-{\displaystyle \int \frac{d}{dx}{\left(logx\right)}^2\cdot {\displaystyle \int xdxdx}}}\\ ={\left(logx\right)}^2*\frac{x^2}{2}-{\displaystyle \int 2logx*\frac{1}{2}*\frac{x^2}{2}dx}\\ =\frac{x^2}{2}{\left(logx\right)}^2-{\displaystyle \int logx\cdot xdx}\end{array}$

$\begin{array}{l}=\frac{x^2}{2}{\left(logx\right)}^2-\left[logx{\displaystyle \int xdx-{\displaystyle \int \frac{d}{dx}logx{\displaystyle \int xdxdx}}}\right]\\ =\frac{x^2}{2}{\left(logx\right)}^2-\frac{x^2}{2}logx+{\displaystyle \int \frac{x}{2}dx}\\ =\frac{x^2}{2}{\left(logx\right)}^2-\frac{x^2}{2}logx+\frac{x^2}{4}+\; C\end{array}$

$\begin{array}{l}{\displaystyle \int ta{n}^{-1}xdx={\displaystyle \int \left(ta{n}^{-1}x\right)1\cdot dx}}.\\ =ta{n}^{-1}x{\displaystyle \int dx-{\displaystyle \int \frac{d}{dx}ta{n}^{-1}x{\displaystyle \int dxdx}}}\\ =xta{n}^{-1}x-{\displaystyle \int \frac{1}{1+x^2}\cdot xdx}.\\ =xta{n}^{-1}x-\frac{1}{2}{\displaystyle \int \frac{2x}{1+x^2}dx}\\ =xta{n}^{-1}x-\frac{1}{2}\cdot log\left|1+x^2\right|+\; C\end{array}$

$\begin{array}{l}{\displaystyle \int xse{c}^{2}xdx=x{\displaystyle \int se{c}^{2}xdx-{\displaystyle \int \frac{dx}{dx}{\displaystyle \int se{c}^{2}xdx}dx.}}}\\ =xtanx-{\displaystyle \int tanxdx}\\ =xtanx-\left(-log\left|cosx\right|\right)+\; C\\ =xtanx+log\left|cosx\right|+\; C\end{array}$

Kindly go through the solution

$LetI={{\displaystyle \int }}^{\text{?}}{\left(si{n}^{-1}x\right)}^{2}dx$

Putting sin^-1x =θ=> x = sinθ, dx = cosθdθ.

$So,I={{\displaystyle \int }}^{\text{?}}{\theta }^2\cdot cos\theta d\theta ={\theta }^2{{\displaystyle \int }}^{\text{?}}cos\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d}{d\theta }{\theta }^2{{\displaystyle \int }}^{\text{?}}cos\theta d\theta d\theta .$

$={\theta }^{2}sin\theta -{{\displaystyle \int }}^{\text{?}}2\theta sin\theta d\theta .\\ ={\theta }^{2}sin\theta -2{{\displaystyle \int }}^{\text{?}}\theta sin\theta d\theta .$

$\begin{array}{cc}\hfill \hfill & \hfill ={\theta }^{2}sin\theta -2\left[\theta {{\displaystyle \int }}^{\text{?}}sin\theta d\theta -{{\displaystyle \int }}^{\text{?}}\frac{d\theta }{d\theta }{{\displaystyle \int }}^{\text{?}}sin\theta d\theta d\theta \right]\hfill \end{array}\\$

$={\theta }^{2}sin\theta -2\left[\theta (-cos\theta )-{{\displaystyle \int }}^{\text{?}}(-cos\theta )d\theta \right]\\$

$={\theta }^{2}sin\theta +2\theta cos\theta -2sin\theta +C$

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

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}$