mathematics-class-12-unit-07-chapter-14-indefinite-integral-l-7-7-5s13z8heovu

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ Denominator =1-2 \sin ^2 x \cos ^2 x $
- $ =1-2\left(1-\cos ^2 x\right) \cos ^2 x $    
- $ =\left(1-2 \cos ^2 x+2 \cos ^4 x\right) $    
- $ I=\int\left(1-2 \cos ^2 x\right) d x $

- $ [\cos 2 x \\ =2 \cos ^2 x-1]  [\cos ^2 x=\frac{1+\cos 2 x}{2}] $

- = $ x-2 [\int \frac{1+\cos 2 x}{2} d x]  $

- = $ x-x-\frac{\sin 2 x}{2}+c $

- = $ -\sin x \cos x+c $

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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"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =\int \frac{d t}{a+(a+b) t^2} $

- $ =\frac{1}{(a+b)} \int \frac{d t}{\left(\frac{a}{a+b}\right)+t^2} $

- $ =\frac{1}{a+b} \int \frac{d t}{\alpha^2+t^2} \quad [\alpha=\sqrt{a / a+b}] $

- $ =\frac{1}{a+b} \times \frac{1}{\alpha} \cdot \tan ^{-1} \frac{t}{\alpha}+c $

- $ =\frac{1}{a+b} \times \frac{\sqrt{a+b}}{\sqrt{a}} \tan ^{-1} \frac{t \sqrt{a+b}}{\sqrt{a}}+c $

- $ =\frac{1}{\sqrt{a(a+b)}} \cdot \tan ^{-1} \frac{t \sqrt{a+b}}{\sqrt{a}}+c $

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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>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-5 </primary>
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- $I=\int \underset{II}1.\underset{I}\sin ^{-1} \frac{2 x}{1+x^2} d x$

- Solution:
- $[\frac{d}{d x} \sin ^{-1} x=\frac{1}{\sqrt{1-x^2}}]$

- $ =\sin ^{-1} \frac{2 x}{1+x^2} \cdot x-\int \frac{1}{\sqrt{1-\left(\frac{2 x}{1+x^2}\right)^2}} \times \frac{d}{d x}\left(\frac{2 x}{1+x^2}\right) \times x d x $

- $ \left[\frac{d}{d x}(\frac{2 x}{1+x^2}) = \frac{\left(1+x^2\right) \times 2-2 {x} \times 2 {x}}{\left(1+x^2\right)^2} =\frac{2\left(1-x^2\right)}{\left(1+x^2\right)^2}\right] $

- $ \sqrt{1-\left(\frac{2 x}{1+x^2}\right)^2} = \sqrt{1-\frac{4 x^2}{\left(1+x^2\right)^2}}=\sqrt{\frac{\left(1+x^2\right)^2-4 x^2}{\left(1+x^2\right)^2}} $

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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution  $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Problem-6 </footer>

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

- $ \Rightarrow I =x \sin ^{-1} \frac{2 x}{1+x^2}-\int \frac{x}{\frac{1-x^2}{(1+x^2)}} \times \frac{2(1-x^2)}{(1+x^2)^2} d x $

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

- $ =x \sin ^{-1} \frac{2 x}{1+x^2}- \log \left|1+x^2\right|+c $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-6 $\rightarrow$ Problem-7 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-6 </primary>
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- $I=\int \cos ^{-1}\left(2 x^2-1\right) d x$

- Solution:
- $ x =\cos \theta \Rightarrow d x=-\sin \theta d \theta $

- $ I =\int \cos ^{-1}\left(2 \cos ^2 \theta-1\right) (-\sin \theta) d \theta $    
- $ =\int 2 \cdot \theta(-\sin \theta) d \theta $   
- $=-2 \int \theta \sin \theta d \theta$

- $=-2\left[\theta(-\cos \theta)-\int 1 \cdot(-\cos \theta) d \theta\right] $  
- $ =+2 \theta \cos \theta-2 \sin \theta+c $     
- $ =2 x \cos ^{-1} x-2 \sqrt{1-x^2}+c $

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<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Problem-6 </span> $\rightarrow$ Problem-7 $\rightarrow$ Problem-8 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-7 </primary>
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- $I=\int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x$

- Solution:
- $[ \cos 2 \theta=1-2 \sin ^2 \theta =2 \cos ^2 \theta-1 ]$

- [$ x=\cos 2 \theta $

- $ d x=-2 \sin 2 \theta d \theta $]

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

- $ I=\int \tan ^{-1}(\tan \theta) \times(-2 \sin 2 \theta) d \theta $    
- $ =-2 \int \underset{I}{\theta}  \underset{II}{\sin 2 \theta} d \theta $

- $\theta \rightarrow x$

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<footer style = "font-size: 18px">Problem-5 $\rightarrow$ Problem-6 $\rightarrow$ <span style = "color:blue"> Problem-7 </span> $\rightarrow$ Problem-8 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-8 </primary>
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- $\quad I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$

- Solution:
- [$ \sqrt{x}=\cos 2 \theta $    
- $\therefore \left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)=\frac{1-\cos 2 \theta}{1+\cos 2 \theta}=\frac{\sin ^2 \theta}{\cos ^2 \theta}, $ ]  [$ \frac{1}{2 \sqrt{x}} d x$
- $=2 \sin 2 \theta d \theta $    
- $ d x=-4 \cos 2 \theta \sin 2 \theta d \theta $    
- $ =-2 \sin 4 \theta d \theta $  ]
- $ I =\int \tan \theta \times(-2 \sin 4 \theta) d \theta $    
- $ =-2 \int \frac{\sin \theta}{\cos \theta} \times 2 \sin 2 \theta \cos 2 \theta d \theta $    
- $ =-4 \int \frac{\sin \theta}{\cos \theta} \times 2 \sin \theta \cos \theta \cdot \cos 2 \theta d \theta $

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<footer style = "font-size: 18px">Problem-6 $\rightarrow$ Problem-7 $\rightarrow$ <span style = "color:blue"> Problem-8 </span> $\rightarrow$ Solution $\rightarrow$ pronlem -9</footer>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =-8 \int \sin ^2 \theta \cos ^2 \theta d \theta, $    
- $ =-2 \int\left(4 \sin ^2 \theta \cos ^2 \theta\right) d \theta $    
- $ =-2 \int \sin ^2 2 \theta d \theta $    
- $ [\cos 2 \theta=1-2 \sin ^2 \theta] $   
- $ [\sin ^2 2 \theta=\frac{1-\operatorname{cos 4} \theta}{2}] $    
- $ =-2 \int\left(\frac{1-\cos 4 \theta}{2}\right) d \theta $    
- $ =-\theta+\frac{\sin 4 \theta}{4}+c $    
- ($ \sqrt{x}=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} \sqrt{x} $  
- ($ \sin 2\theta  =\sqrt{1-\cos^2 2\theta} =\sqrt{1-x} $ )

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<footer style = "font-size: 18px">Problem-7 $\rightarrow$ Problem-8 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-9 $\rightarrow$ Problem-10 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Problem-9 </primary>
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- $ I=\int e^x({f(x)}+f^{\prime}(x)) d x $

- Solution:
- $ =\int \underset{I_1}{e^x f(x)} d x+\int \underset{I_2}{e^x f^{\prime}}(x) d x $

- $ I_1 =\int \underset{II}{e^x} \underset{I}{f(x)} d x$

- $ =f(x) e^x- \underset{I_2}{\int f^{\prime}(x) e^x} d x+c $

- $ I_1 =f(x) \cdot e^x-I_2 + c $

- $ \Rightarrow I =f(x) \cdot e^x-I_2+I_2+c $

- $ \Rightarrow I=e^x f(x)+c $

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<footer style = "font-size: 18px">Problem-8 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-9 </span> $\rightarrow$ Problem-10 $\rightarrow$ Problem-11 </footer>

### <Success style="font-size:25px"> Indefinite Integral L-7 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ =\int e^t \log t d t+\int \frac{e^t}{t^2} d t $

- $ =\log t . e^t-\int \frac{1}{t}. e^t d t+\int \frac{e^t}{t^2} d t $

- $ =e^t \log t-\left(\int e^t\left(\frac{1}{t}-\frac{1}{t^2}\right) d t\right) $

- $ =e^t \log t-e^t \cdot \frac{1}{t}+c $

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

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<footer style = "font-size: 18px">Problem-10 $\rightarrow$ Problem-11 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-12 </footer>

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

- $ =\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2} $

- [Here,   $\tan \frac{x}{2} = f(x)$ ; $\frac{1}{2} \sec ^2 \frac{x}{2} = f^{\prime} (x)$]

- $\Rightarrow  I=\int e^x\left(\tan \frac{x}{2}+\frac{1}{2} \sec ^2 \frac{x}{2}\right) d x $

- $ I=e^x \tan \frac{x}{2}+c $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-12 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-13 $\rightarrow$ Solution </footer>