mathematics-class-12-unit-08-chapter-01-integral-calculus-l-1-6-ftayk7ubong

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- If $I=\frac{2}{\pi} \int_{\pi / 4}^{\pi / 4} \frac{d x}{\left(1+e^{\sin x}\right)(2-\cos 2 x)}$ then $27 I^2=$ ?

- Solution:

- Recall: $\int_{-a}^a f(x) d x=\int_0^a[f(x)+f(-x)] d x$

- Pf: $\int_{-a}^a f(x) d x=\int_{-a}^0 f(x) d x+\int_0^a f(x) d x$
- Now putting $x=-y$ in the first integral, $d x=-d y$ and
- when $x=-a, y=a$;
- when $x=0, y_0=0$

- So, $\int_{-a}^0 f(x) d x=\int_a^x f(-y)(-d y)$

- $ -\int_a^a f(x) d y-\int_0^a(1-y) d y $

- $ =\int_0^a f(-x) d x $

- i.e, $\int_{-a}^a f(x) d x =\int_0^a[f(-x)+f(-x)] d x$

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<footer style = "font-size: 18px"> $\rightarrow$ Integral calculus lecture-1 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \Rightarrow f(x)+f(-x)  =\frac{(1+e^{\sin x})}{\left(1+e^{\sin x}\right)(2-\cos 2 x)} $

- $ =\frac{1}{2-\cos 2 x} $

- $=\frac{1}{2-\left(2 \cos ^2 x-1\right)}$

- $ =\frac{1}{2-\cos 2 x} $

- $=\frac{1}{2-\left(2 \cos ^2 x-1\right)} $

- $=\frac{1}{3-2 \cos ^2 x}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \therefore I  =\frac{2}{\pi} \int_0^{\pi / 4} \frac{1}{(3-2 \cos ^2 x)} d x $

- $ =\frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{3 \sec ^2 x-2} d x$

- $ =\frac{2}{\pi} \int_0^{\pi / 4} \frac{\sec ^2 x}{3(1+\tan ^2 x)-2} d x$

- Put $u=\tan x$. Then $d u=\sec ^2 x d x$ When $x=0, u=0$

- When $x=\frac{\pi}{4}, u=\tan \frac{\pi}{4}=1$

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- So, $\begin{aligned} I & =\frac{2}{\pi} \int_0^1 \frac{d u}{3 u^2+1} \\ & =\frac{2}{3 \pi} \int_0^1 \frac{d u}{u^2+\left(\frac{1}{\sqrt{3}}\right)^2}\end{aligned}$

- $=\left.\frac{2}{3 \pi} \frac{1}{(1 / \sqrt{3})} \tan ^{-1}\left(\frac{u}{1 / \sqrt{3}}\right)\right|_0 ^1$

- $ =\frac{2}{\sqrt{3} \pi}\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} 0\right) $

- $=\frac{2}{\sqrt{3} \pi} \times \frac{\pi}{3}=\frac{2}{3 \sqrt{3}} $

- $ \Rightarrow \quad I^2  =\frac{4}{27} \Rightarrow 27 I^2=4$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore \quad I=\int_0^{\pi / 2} \frac{3 \sqrt{\sin \varphi}}{\left(\sqrt{\sin \varphi}+\sqrt{\cos \varphi)^5}\right.} d \varphi$ $ =\int_0^{\pi / 2} \frac{3 \sqrt{\sin \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$

- Adding (i) \& (ii), we get

- $2 I=\int_0^{\pi / 2} \frac{3(\sqrt{\cos \theta}+\sqrt{\sin \theta})}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^5} d \theta$
- $ =\int_0^{\pi / 2} \frac{3}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^4} d \theta $
- $ =\int_0^{\pi / 2} \frac{3}{\cos ^2 \theta(1+\sqrt{\tan \theta})^4} d \theta$
- $=\int_0^{\pi / 2} \frac{3 \sec \theta}{(1+\sqrt{\tan \theta})^4} d \theta$
- $ =\int_0^{\pi / 2} \frac{3}{(\sqrt{\cos \theta}+\sqrt{\sin -\theta})^4} d \theta $
- $=\int_0^{\pi / 2} \frac{3}{\cos ^2 \theta(1+\sqrt{\tan \theta})^4} d \theta$
- $=\int_0^{\pi / 2} \frac{3 \sec^2 \theta}{(1+\sqrt{\tan \theta})^4} d \theta$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \text { When } \theta=0, t=0$

- $ \text { When } \theta = \frac{\pi}{2} , t=0$

- $ =3 \int_0^{\infty} \frac{(t+1)-1}{(t+1)^4} d t $ 
- $ =3\left[\int_0^{\infty}(t+1)^{-3} d t-\int_0^{\infty}(t+1)^{-4} d t\right] $
- $=\left.3\left(\frac{-1}{2(t+1)^2}+\frac{1}{3(t+1)^3}\right)\right|_0 ^{\infty}$
- $ =3\left[\int_0^{\infty}(t+1)^{-3} d t-\int_0^{\infty}(t+1)^{-4} d t\right] $
- $ =\left.3\left(\frac{-1}{2(t+1)^2}+\frac{1}{3(t+1)^3}\right)\right|_0 ^{\infty} $
- $=3\left(0+\frac{1}{2}-\frac{1}{3}\right)=3 \times \frac{1}{6}=\frac{1}{2} $

- $\therefore I=\frac{1}{2}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- Find the value of the integral
- $I=\int_0^{1 / 4} \frac{1+\sqrt{3}}{\left[(x+1)^2(1-x)^6\right]^{1 / 4}} d x$

- Solution:

- $I=(1+\sqrt{3}) \int_0^{1 / 2} \frac{d x}{(x+1)^{1 / 2}(1-x)^{3 / 2}}$

- $ =(1+\sqrt{3}) \int_0^{1 / 2} \frac{d x}{(1-x) \sqrt{1-x^2}}$

- Put $x=\sin \theta$. The $d x=\cos \theta d \theta$ When $x=0, \theta=0$

- When $x=\frac{1}{2}, \theta=\frac{\pi}{6}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \therefore I  =(\sqrt{3}+1) \int_0^{\pi / 6} \frac{\cos \theta d \theta}{(1-\sin \theta) \cos \theta} $

- $=(\sqrt{3}+1) \int_0^{\pi / 6} \frac{1+\sin \theta}{\cos ^2 \theta} d \theta$

- $ =(\sqrt{3}+1) \int_0^{\pi /6 } {(sec ^2 \theta+\sec \theta \tan \theta) d \theta} $

- $ =\left.(\sqrt{3}+1)(\tan \theta+\sec \theta)\right|_0 ^{\pi / 6} $

- $=(\sqrt{3}+1)\left[\frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}-0-1\right]$

- $ =(\sqrt{3}+1)\left(\frac{3}{\sqrt{3}}-1\right) $

- $ =(\sqrt{3}+1)(\sqrt{3}-1)=2 $

- $\therefore I  =2$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0)=0, f\left(\frac{\pi}{2}\right)=3$ and $f^{\prime}(0)={1}$.

- If $g(x)=\int_x^{\pi / 2}\left[f^{\prime}(t) \operatorname{cosec} t-\cot t \operatorname{cosec} t f(t)\right] d t$, for $x \in\left(0, \frac{\pi}{2}\right]$, the
- $\lim _{x \rightarrow 0} g(x)=\text { ? }$

- Soln: Note that $f^{\prime}(t) \operatorname{cosec} t-\operatorname{cosec} t \cot t f(t)$
- $ =\frac{d}{d t}[f(t) \text { cosect }]$
- 1) $g(x)=\int_x^{\pi / 2} \frac{d}{d t}[f(t) \operatorname{cosec} t] d t$

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\begin{aligned}& =\left.f(t) \operatorname{cosec} t\right|_x ^{\pi / 2} \\ =f\left(\frac{\pi}{2}\right) \operatorname{cosec} \frac{\pi}{2}-f(x) \operatorname{cosec} x \\ =3-f(x) \operatorname{cosec} x\end{aligned}$

- Note that cosec 0  is not defined, so we connot find $g(0)$

- But $\lim _{x \rightarrow 0} g(x)=3-\lim _{x \rightarrow 0} \frac{f(x)}{\sin x}\left(\frac{0}{0}\right)$
- $ =3-\lim _{x \rightarrow 0} \frac{f(x) / x}{\sin x / x}$

- $=3-\lim _{x \rightarrow 0} \frac{f(x) / x}{\sin x / x} $ $ =3-\lim _{x \rightarrow 0} \frac{f(x)}{x} \quad\left(\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right) $ $ =3-\lim _{x \rightarrow 0} \frac{f(x)-f(0)}{x-0} \quad(\because f(0)=0) $ $ =3-f^{\prime}(0)=3-1=2$

- Ans: $\lim _{x \rightarrow 0} g(x)=2$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

<h3 id="success-stylefont-size25px-integral-calculus-l-1-success"><Success style="font-size:25px"> Integral Calculus L-1 </Success></h3>
<h3 id="primary-stylefont-size24px-solution-primary"><primary style="font-size:24px"> Solution </primary></h3>
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<p>Note that $\ln 99=\int_1^{99} \frac{1}{x} d x=\sum_{k=1}^{98} \int_k^{k+1} \frac{1}{x} d x$</p>
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<p>Now, if $k<x<k+1$, the $k+1<x+1$</p>
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<p>$\begin{gathered}\Rightarrow \frac{k+1}{x+1}<1 \\ \therefore \frac{k+1}{x(x+1)}<\frac{1}{x+1}\end{gathered}$</p>
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<p>So, $\quad \int_k^{k n} \frac{k n}{x(x+1)} d x<\int_k^{x+1} \frac{1}{x} d x$</p>
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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\text { 1) } I=\sum_{h=1}^{98} \int_h^{h+1} \frac{k+1}{x(x+1)} d x<\ln 99$

- $\therefore(A)$ is false; (B) is true.

- Similarly, $\frac{k+1}{x}>1$ if $k<x<k+1$ i. $\frac{k+1}{x(x+1)}>\frac{1}{x+1}$

- $\int_{k+1}^{k+1} \frac{k+1}{x(x+1)} d x>  \int_k^{k+1} \frac{1}{x+1} d x $

- $ \therefore=\sum_{k=1}^{98} \int_k^{k+1} \frac{k+1}{x(x+1)} d x  >\int_1^{99} \frac{1}{x+1} d x $ $=\left.\ln (x+1)\right|_1 ^{99} $ $=\ln 100-\ln 2 $ $=\ln 50$

- Clearly, $\ln 50>1>\frac{49}{50}, \operatorname{sin} I>\frac{49}{50}$
- $\therefore$ (D) is true; (C) is false

- Remark, Note that $\frac{49}{50}=\frac{98}{100}$
- So if we show that $\int_h^{\frac{k+1}{50}} \frac{k+1}{x(x+1)} d x>\frac{1}{100}$ for each $k$, the $I>\frac{98}{100}=\frac{49}{50}$

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Put $x=k+y$. When $x=k, y=0$ the $x=k+1, y=1$

- $\int_k^{k+1} \frac{k+1}{x(x+1)} d x  =\int_0^1 \frac{k+1}{(k+y)(k+1+y)} d y $

- $ >\int_0^1 \frac{1}{(k+1+y)} d y \quad\left(\because \frac{k+1}{k+y}>1\right) $

- $ >\frac{1}{k+2} \geqslant \frac{1}{100} \quad \text { if } 1 \leqslant k \leqslant 98 $

- $\therefore \quad \int_k^{k+1} \frac{k+1}{x(x+1)} d x >\frac{1}{100}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $[x]=$ a greatest integer less than a equal to $x$.

- $=n$, if $n \leqslant x<n+1$ for any integer $n$

- let $g(x)=\frac{x f\left(x^2\right)}{2+f(x+1)}$

- Now  :

- $f\left(x^2\right)  = \begin{cases}{[x^2]} & \text { if } x^2 \leq 2 \\\0 & \text { if } x^2>2\end{cases} $
- $= \begin{cases}0 & \text { if }-1<x<1 \\\ 1 & \text { if } 1 \leq x<\sqrt{2} \\\ 0 & \text { if } x>\sqrt{2}\end{cases} $
- $ \text { Also, } f(x+1)  =\begin{cases}{[x+1] \qquad \text { if } x+1 \leq 2} \\\ 0 \qquad \qquad \text { if } x+1>2\end{cases}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $=\begin{cases}{[x+1] \qquad \text { if } -1 \leq x \leq 1} \\\ 0 \qquad \qquad \text { if } x>1\end{cases}$

- $=\begin{cases}0 & \text { if }-1\leq x<0 \\\ 1 & \text { if } 0 \leq x<1 \\\ 0 & \text { if } x>1 \end{cases} $

- $g\left(x\right)  = \begin{cases}{\frac{(x)(1)}{2+0}} & \text { if } 1 \leq x < \sqrt{2} \\\0 & \text { if } x>\sqrt{2}\end{cases} $

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Integral Calculus L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $= \begin{cases}\frac{x}{2} & \text { if } 1 \leq x<\sqrt{2} \\\ 0 & \text { otherwise }\end{cases} $

- $ \therefore I=\int_{-1}^2 g(x) d x $

- $ =\int_{-1}^1 g(x) d x+\int_{1}^{\sqrt{2}} g(x) d x +\int_2^2 g(x) d x$

- $\begin{aligned}=\int_1^{\sqrt{2}} \frac{x}{2} d x=\left.\frac{x^2}{4}\right|_1 ^{\sqrt{2}} & =\frac{1}{4}(2-1) \\& =\frac{1}{4}\end{aligned}$

- So, $I=\frac{1}{4}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>