Maths Integrals

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int e^{ta{n}^{-1}x}}\left(\frac{1+x+x^2}{1+x^2}\right)dx\\ Putta{n}^{-1}x=t\Rightarrow \frac{1}{1+x^2}.dx=dt\\ ={\displaystyle \int e^t}\left(1+tant+ta{n}^{2}t\right)dt={\displaystyle \int e^t\left(se{c}^{2}t+tant\right)dt}\\ Here,f\left(t\right)=tant\\ \therefore f^{\text{'}}\left(t\right)=se{c}^{2}t\\ =e^t.f\left(t\right)=e^{t}tant=e^{ta{n}^{-1}x}.x+C\\ \left[?{\displaystyle \int e^t}\left[f\left(x\right)+f^{\text{'}}\left(x\right)\right]dx=e^{x}f\left(x\right)+C\right]\\ Hence,I=e^{ta{n}^{-1}x}.x+C\end{array}$

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

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

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

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx\\ Put{x}^2=tforthepurposeofpartialfraction.\\ Weget\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}\\ Put\frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A}{t+a^2}+\frac{B}{t+b^2}\left[whereAandBarearbitrar\mathrm{yconstants}.\right]\\ \frac{t}{\left(t+a^2\right)\left(t+b^2\right)}=\frac{A\left(t+b^2\right)+B\left(t+a^2\right)}{\left(t+a^2\right)\left(t+b^2\right)}\\ \Rightarrow t=At+A{b}^2+Bt+B{a}^2\\ Comparingtheliketerms,weget\\ A+B=1andA{b}^2+B{a}^2=0\\ \Rightarrow A=\frac{-a^2}{b^2}B\\ \therefore \frac{-a^2}{b^2}B+B=1\\ B\left(\frac{-a^2}{b^2}+1\right)=1\Rightarrow B\left(\frac{-a^2+b^2}{b^2}\right)=1\\ \Rightarrow B=\frac{b^2}{b^2-a^2}andA=\frac{-a^2}{b^2}*\frac{b^2}{b^2-a^2}=\frac{a^2}{a^2-b^2}\\ So,A=\frac{a^2}{a^2-b^2}andB=\frac{-b^2}{a^2-b^2}\\ \therefore {\displaystyle \int \frac{x^2}{\left(x^2+a^2\right)\left(x^2+b^2\right)}}dx=\frac{a^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+a^2}dx-}\frac{b^2}{a^2-b^2}{\displaystyle \int \frac{1}{x^2+b^2}dx}\\ =\frac{a^2}{a^2-b^2}*\frac{1}{a}ta{n}^{-1}\frac{x}{a}-\frac{b^2}{a^2-b^2}.\frac{1}{b}.ta{n}^{-1}\frac{x}{b}\\ =\frac{a}{a^2-b^2}ta{n}^{-1}\frac{x}{a}-\frac{b}{a^2-b^2}ta{n}^{-1}\frac{x}{b}+C\\ Hence,I=\frac{1}{a^2-b^2}\left[ata{n}^{-1}\frac{x}{a}-bta{n}^{-1}\frac{x}{b}\right]+C.\end{array}$

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

Consider, $I={\displaystyle {\int }_0^{1}ta{n}^{-1}\left(\frac{2x-1}{1+x-x^2}\right)}dx$

$\begin{array}{l}\Rightarrow I={\displaystyle {\int }_0^{1}ta{n}^{-1}\left(\frac{x-\left(1-x\right)}{1+x\left(1-x\right)}\right)}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}x-ta{n}^{-1}\left(1-x\right)\right]}dx-----(1)\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(1-1+x\right)\right]}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}x\right]}dx\\ \Rightarrow I={\displaystyle {\int }_0^1\left[ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(x\right)\right]}dx-----(2)\end{array}$

Adding (1) and (2), we get

$\begin{array}{l}\Rightarrow 2I={\displaystyle {\int }_0^1\left[ta{n}^{-1}x-ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}\left(1-x\right)-ta{n}^{-1}x\right]}dx\\ \Rightarrow 2I=0\\ \Rightarrow I=0\end{array}$

Thus, the correct option is B.

Let $=\int \frac{cos2xdx}{{(sinx+cosx)}^2}$

$=\int \frac{co{s}^{2}x-si{n}^{2}x}{{(sinx+cosx)}^2}dx\{?cos2x=co{s}^{2}x-si{n}^{2}x$

$=\int \frac{(cosx+sinx)(cosx-sinx)}{{(sinx+cosx)}^2}dx\{?a^2-b^2=(x+b)(x-b)$

$=\int \frac{(cosx-sinx)}{sinx+cosx.}dx$

$=log|sinx+cosx|+c\{{\displaystyle \int \frac{f^{\prime }(x)}{f(x)}dx=log|f(x)|+c}$

So, option B is correct.

Let $=\int \frac{dx}{e^x+e^{-x}}$

$=\int \frac{dx}{e^x+\frac{1}{e^x.}}=\int \frac{e^{x}dx}{e^x·e^x+1}$

$=\int \frac{e^{x}dx}{e^{2x}+1}$

Putting e^x = t

e^xdx = dt.

$\therefore =\int \frac{dt}{t^2+1.}$

= tan^{- 1} t + c

= tan^{- 1} (e^x) + c

therefore, Option A is correct.

Let $I={\displaystyle {\int }_0^1{e}^{2-3x}dx}$

We know that,

${\displaystyle {\int }_a^{b}f\left(x\right)}dx=\left(b-a\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(a\right)+f\left(a+h\right)+...+f\left(a\left(n-1\right)h\right)\right]$

Where, $h=\frac{b-a}{n}$

Here, $a=0,b=1$ and $f\left(x\right)e^{2-3x}$

$\Rightarrow h=\frac{1-0}{n}=\frac{1}{n}$

$\therefore {\displaystyle {\int }_0^1{e}^{2-3x}dx}=\left(1-0\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[f\left(0\right)+f\left(0+h\right)+...+f\left(0+\left(n-1\right)h\right)\right]$

$\begin{array}{l}=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2+e^{2-3x}+...+e^{2-3\left(n-1\right)h}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{1+e^{-3h}+e^{-6h}+e^{-9h}+...+e^{-3\left(n-1\right)h}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-{\left(e^{-3h}\right)}^n}{1-\left(e^{-3h}\right)}\right\}\right]=\underset{n\to \infty }{lim}\frac{1}{n}\left[e^2\left\{\frac{1-e^{\frac{-3}{n}n}}{1-e^{\frac{-3}{n}}}\right\}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\\ =\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{e^2\left(1-e^{-3}\right)}{1-e^{\frac{-3}{n}}}\right]=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\frac{1}{n}\left[\frac{1}{e^{\frac{-3}{n}}-1}\right]\end{array}$

$\begin{array}{l}=e^2\left(e^{-3}-1\right)\underset{n\to \infty }{lim}\left(-\frac{1}{3}\right)\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]=\frac{e^2\left(e^{-3}-1\right)}{3}\underset{n\to \infty }{lim}\left[\frac{-\frac{3}{n}}{e^{\frac{-3}{n}}-1}\right]\\ =\frac{-e^2\left(e^{-3}-1\right)}{3}(1)\\ =\frac{-e^{-1}+e^2}{3}\\ =\frac{1}{3}\left(e^2-\frac{1}{e}\right)\end{array}$

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