mathematics-class-12-unit-07-chapter-07-definite-integral-l-7-7-z9yju6covpw

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Recall </primary>
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- $f(x)$ continuous on $[a, b], f(x)>0$

- $ m(b-a) \leqslant \int_a^b f(x) d x \leqslant {M(b-a)} $

- $ m \leqslant f(a) \leqslant M \quad\forall x[a, b] $

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<footer style = "font-size: 18px"> $\rightarrow$ Definite integral lecture-7 $\rightarrow$ <span style = "color:blue"> Recall </span> $\rightarrow$ Remark $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- <ins>**Evaluate**</ins>

- $ \int_{-1 / 2}^{1 / 2}[[x]+\log(\frac{1+x}{1-x})] d x $

- $ =\int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log(\frac{1+x}{1-x}) d x $

- $ \log (\frac{1+x}{1-x})=h(x) $

- $ h(-x)=\log(\frac{1-x}{1+x}) $ $ =-\log (\frac{1+x}{1-x}) $  $ =-h(x) $

- $\therefore \int_{-1 / 2}^{1 / 2} \log(\frac{1+x}{1-x}) d x = 0$

- $\therefore \int_{-1 / 2}^{1 / 2}[x] d x+\int_{-1 / 2}^{1 / 2} \log(\frac{1+x}{1-x}) d x =\int_{-1 / 2}^{1 / 2}[x] d x $

- $ =\int_{-1 / 2}^{0} (-1) d x $  $ +\int_{0}^{1 / 2} (0) d x $  $ = -x|_{-1 / 2} ^0=-\frac{1}{2} $

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<footer style = "font-size: 18px">Recall $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- <ins>**Evaluate**</ins>

- $ I=\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, \quad a>0 $

- $ x=-t \quad d x=-d t \quad I=\int_{-\pi}^{-\pi} \frac{-\cos ^2 (-t) d t}{1+a^{-t}} $

- $ I=\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x \quad  ......(1) =\int_ {-\pi}^\pi \frac{\cos ^2 t d t}{1+{a}^{-t}}  =\int_{-\pi}^\pi \frac{a^x\cos ^2 x dx}{1+a^x} \quad ......(2) $

- $ \text { (1) +(2) } $

- $ 2 I=\int_{-\pi}^\pi \frac{(1+a^{x}) \cos ^2 x}{(1+a^{x})} d x $  $ 2 I=\int_{-\pi}^\pi \cos ^2 x d x  =2 \int_0^\pi \frac{(1+\cos 2 x)}{2} d x =\pi \Rightarrow I=\frac{\pi}{2} $

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<footer style = "font-size: 18px">Remark $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Problem-3 $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- <ins>**Compute**</ins>

- $ I=\int_ {\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}} =\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan (\frac{\pi}{2}-x)}} $

- $ \int_a^b f(x) d x=\int_a^{b} f(a+b-x) d x $

- $ I=\int_ {\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}} =\int_{\pi / 6}^{\pi / 3} \frac{1+\sqrt{\tan x}-1}{1+\sqrt{\tan x}}) d x $

- $ I=\int_{\pi / 6}^{\pi / 3} 1 d x-\int_{\pi / 6}^{\pi / 3} \frac{1}{1+\sqrt{\tan x}} d x $

- $ 2 I=\frac{\pi}{3}-\frac{\pi}{6} \Rightarrow I = \frac{\pi}{12} $

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ 2 I=\int_0^{\pi / 4}[\log (\frac{2}{1+\tan x})+ log(1+\tan x)] d x $

- $2 I=\int_0^{\pi / 4} \log [\frac{2}{1+\tan x} \times (1+\tan x)] d x $

- $2 I=\int_0^{\pi / 4} \log 2 d x =\log 2 \ \frac{\pi}{4} $

- $ I=\left(\frac{\pi}{8}\right) \log 2 $

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

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- $ I=\int_{-\pi}^\pi \frac{2 x(1+\operatorname{sin} x)}{(1+\cos ^2 x)} d x  $ Compute $I$.

- solution:
- $ I =\int_{-\pi}^{\pi} \frac{2 x d x}{1+\cos ^2 x}+\int_{-\pi}^{\pi} \frac{2 x \sin x}{1+\cos^2 x} d x $

- $ f(x)=\frac{2 x}{1+\cos ^2 x} $

- $ f(-x)=\frac{-2 x}{1+\cos ^2(-x)} $  $ =\frac{-2 x}{1+\cos ^2 x} = -f(x) $

- $ \int_{-a}^a f(x) d x=0 $

- $ [f(x) \text { is odd }] $

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

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ I =\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x $

- $ g(x)=\frac{2x \sin x}{1+\cos ^2x} $
- $ g(-x)=\frac{2(-x) \sin (-x)}{1+\cos ^2(-x)} = \frac{2 x \sin x}{1+\cos ^2 x} $

- $ \quad =g(x) $

- $ \int_{-a}^a g(x) d x=2 \int_0^a g(x) d x $

- $ \int_0^a f(x) d x =\int _0^a f(a-x) d x $

- $ I =2 \int_0^{\pi} \frac{2 x \sin x}{1+\cos ^2 x} \quad ...(1)  =2 \int_0^\pi \frac{2(\pi-x) \sin (\pi-x) d x}{1+(\cos ^2(\pi-x))} $

- $ I =4 \int_0^\pi \frac{(\pi- x) \sin x d x}{1+\cos ^2 x}\quad ...(2) $

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

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ 2 I=4 \int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x $

- $ \text { Let } \cos x=t $

- $ - \sin x d x=d t $

- $ I=2 \pi \int_1^{-1} \frac{-d t}{1+t^2} $

- $ =2 \pi \int_{-1}^1 \frac{d t}{1+t^2} $

- $ I=2 \pi \tan ^{-1} t|_{-1} ^1 $

- $ =2 \pi[\frac{\pi}{4}-(-\frac{\pi}{4})] $

- $ I=2 \pi \frac{\pi}{2}=\pi^2 $

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

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- <ins>Evaluate:</ins>

- $ I=\int_0^\pi \frac{x d x}{1+\cos \alpha \sin x}, 0<\alpha<\pi $

- solution:
- $ I=\int_0^\pi \frac{x d x}{1+\cos \alpha \sin x} \quad ...(1)$

- $ \int_0^a f(x) d x =\int _0^a f(a-x) d x $

- $ I=\int_0^\pi \frac{(\pi-x) d x} {1+\cos \alpha \sin (\pi-x)} \quad ...(2) $

- $ 2 I=\int_0^\pi \frac{\pi d x}{1+\cos \alpha \sin x}=\pi \int_0^\pi \frac{d x}{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+ \cos \alpha \ 2\sin \frac{x}{2} \cos \frac{x}{2}} $

- $ 2 I=\pi \int_0^\pi \frac{\sec ^2 \frac{x}{2} d x}{\tan ^2 \frac{x}{2}+1+2 \tan \frac{x}{2} \cos \alpha} $  
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-6  </span> $\rightarrow$ Solution $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \tan \frac{x}{2}=t $  $ \sec ^2 \frac{x}{2} d x = 2 d t $  $ \tan 0 = 0 $  $ \tan \frac{\pi}{2} = {\infty} $

- $ I=\frac{\pi}{2} \int_0^{\infty} \frac{2 d t}{t^2+1+2 t \cos \alpha+\cos ^2 \alpha-\cos ^2 \alpha} $   $ =\pi \int_0^{\infty} \frac{d t}{(t+\cos \alpha)^2+\sin ^2 \alpha} $  $ =\pi  \frac{1}{\sin \alpha} \tan ^{-1}(\frac{t+\cos \alpha}{\sin \alpha}) |_0^{\infty} $

- $ I =\frac{\pi}{\sin \alpha}[\tan ^{-1} \infty -\tan ^{-1} \cot \alpha] , 0<\alpha<\pi $  $ =\frac{\pi}{\sin \alpha}[\frac{\pi}{2}-(\frac{\pi}{2}-\alpha)] $

- $ I =\frac{\pi \alpha}{\sin \alpha}, 0<\alpha<\pi $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-7 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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- $ I=\int_0^{\pi / 2} \frac{x \sin x \cos x d x}{\cos ^4 x+\sin ^4 x} \quad ...(1) $

- Solution:
- $ \int_0^a f(x) d x $

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

- $ =\int_0^{\pi / 2} \frac{(\frac{\pi}{2}-x) \sin (\frac{\pi}{2}-x) \cos (\frac{\pi}{2}-x) d x} {{\cos ^4(\frac{\pi}{2}-x)+\sin ^4(\frac{\pi}{2}-x)}} $

- $ I=\int_0^{\pi / 2} \frac{(\frac{\pi}{2}-x) \cos x \sin x d x}{\sin ^4 x+\cos ^4 x} \quad ...(2) $

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Solution $\rightarrow$ Evaluate </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ (1)+(2) $

- $ 2 I =\int_0^{\pi / 2} \frac{\frac{\pi}{2} \sin x \cos x}{\cos ^4 x+\sin ^4 x} d x $

- $ I =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x $   $ =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\frac{\sin x \cos x}{\cos ^4 x}}{{1+\tan ^4 x}} dx $

- $ I =\frac{\pi}{4} \int_0^{\pi / 2} \frac{\tan x \sec ^2 x d x}{1+\tan ^4 x} $  $ \tan ^2 x=t $

- $ 2 \tan x \sec ^2 x {d x}=d t  $   $ \tan 0=0 $ $ \tan \frac{\pi}{2}=\infty $

- $ I=\frac{\pi}{4} \int_{t=0}^{t=\infty} \frac{\frac{d t}{2}}{1+t^2} $

- $ I=\frac{\pi}{8} \int_0^{\infty} \frac{d t}{1+t^2} =\left.\frac{\pi}{8} \tan ^{-1} t\right|_0 ^{\infty}$

- $ I=\frac{\pi}{8}[\tan ^{-1} \infty-\tan ^{-1} 0] $  $\quad I=\frac{\pi^2}{16} $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Evaluate $\rightarrow$ Evaluate </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Evaluate </primary>
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- $ \int_0^{\pi / 4} \frac{(\sin x+\cos x)}{9+16 \sin 2 x} dx $ 
- $ =\int_0^{\pi / 4} \frac{\sin x+\cos x}{25-16+16 \sin 2 x} d x $
- $ =\int_0^{\pi / 4} \frac{(\sin x+\cos x) d x}{25-16(1-\sin 2 x)} $
- $ =\int_0^{\pi / 4} \frac{(\sin x+\cos x) d x}{25-16(\sin ^2 x+\cos ^2 x-2 \sin x \cos x)} $ 
- $ =\int_0^{\pi / 4} \frac{(\sin x+\cos x) d x}{25-16(\sin x-\cos x)^2} $

- $ \text { Let } \sin x-\cos x=t $  $ (\cos x+\sin x) d x=d t $

- $ x = \frac{\pi}{4} , t = 0 $ $ x = 0 , t = -1 $

- $ =\int_{-1}^0 \frac{d t}{25-16 t^2} $  $ =\frac{1}{16} \int_{-1}^0 \frac{d t}{(\frac{5}{4})^2-t^2} $

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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Evaluate </span> $\rightarrow$ Evaluate $\rightarrow$ Problem-9 </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Evaluate </primary>
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- $ \because \int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log |\frac{a+x}{a-x}| $

- $ I =\frac{1}{16} \frac{1}{2 \times \frac{5}{4}}(\log |\frac{\frac{5}{4}+t}{\frac{5}{4}-t}|)|_{-1} ^0 $ 
- $ =\frac{1}{40}[\log 1-\log |\frac{1 / 4}{3 / 4}|] $ 
- $ =\frac{1}{40}(-1) \log (\frac{1}{9}) $ 
- $ =\frac{1}{40} \log 9 =\frac{1}{40} \log 3^2=\frac{2}{40} \log 3 $

- $ =\frac{1}{20} \log 3 $

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<footer style = "font-size: 18px">Solution  $\rightarrow$ Evaluate $\rightarrow$ <span style = "color:blue"> Evaluate </span> $\rightarrow$ Problem-9 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ x^2=\frac{2}{1+x^2} $

- $ y=\frac{2}{1+y} $

- $ y^2+y-2=0 $

- $ y=\frac{-1 \pm \sqrt{1+8}}{2}=\frac{-1 \pm 3}{2} $  $ y=-2,1 $

- $ y=1=x^2 \\\\ \Rightarrow x= \pm 1 $

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<footer style = "font-size: 18px">Evaluate $\rightarrow$ Problem-9 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-10 </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \text { Required Area }  =\int_{-1}^1 \underbrace{[\frac{2}{1+x^2}-x^2] d x} _{\text { elementary area }} $

- $ =2 \int_{-1}^1 \frac{d x}{1+x^2}-\int_{-1}^1 x^2 d x $

- $ =2 \tan ^{-1} x|_{-1} ^1 $

- $ -\frac{x^3}{3}|_{-1} ^1 $

- $ =2(\frac{\pi}{4}-(-\frac{\pi}{4})-\frac{1}{3}(1-(-1))=\pi-\frac{2}{3} $

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<footer style = "font-size: 18px">Problem-9 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-10 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Definite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-10 </primary>
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- Sketch the curves and identify the region bounded by $x=\frac{1}{2}, x=2, y=\log x$ and $y=2^x$.

- $ \int_{\frac{1}{2}}^2\left[2^x-\log x\right] d x $

- $ =\biggl[\frac{2^x}{\log 2}-(x \log x-x)\biggl] |_{\frac{1}{2}} ^2 $

- $ =\frac{2^2}{\log ^2}-2 \log 2+2-\frac{\sqrt{2}}{\log 2}-\frac{1}{2} \log \frac{1}{2}+\frac{1}{2} $

- $ =\frac{4-\sqrt{2}}{\log 2}+\frac{3}{2}-\frac{5}{2} \log 2  $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-10 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>