Prove the following
Question: $\sin x=\frac{2 t}{1+t^{2}}, \quad \tan y=\frac{2 t}{1-t^{2}}$ Solution: Given, sin x = 2t/(1 + t2), tan y = 2t/ (1 t2) Now, taking $\sin x=\frac{2 t}{1+t^{2}}$ Differentiating both sides w.r.t $t$, we get $\cos x \cdot \frac{d x}{d t}=\frac{\left(1+t^{2}\right) \cdot \frac{d}{d t}(2 t)-2 t \cdot \frac{d}{d t}\left(1+t^{2}\right)}{\left(1+t^{2}\right)^{2}}$ $\cos x \cdot \frac{d x}{d t}=\frac{2\left(1+t^{2}\right)-2 t \cdot 2 t}{\left(1+t^{2}\right)^{2}}$ $\frac{d x}{d t}=\frac{2+2 t^{...
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Question: Evaluate $\int \frac{\sin x}{\cos 2 x} d x$ Solution: $\operatorname{Let} I=\int \frac{\sin x}{\cos 2 x} d x$ We know $\cos 2 x=2 \cos ^{2} x-1$ Putting values in I we get, $I=\int \frac{\sin x}{\cos 2 x} d x=\int \frac{\sin x}{2 \cos ^{2} x-1} d x$ Put $\cos x=t$ Differentiating w.r.t to $x$ we get, $\sin x d x=-d t$ Putting values in integral we get, $I=-\int \frac{d t}{2 t^{2}-1}=-\int \frac{d t}{(\sqrt{2} t)^{2}-(1)^{2}}$ Again put $\sqrt{2} \times \mathrm{t}=\mathrm{u}$ Differenti...
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Question: If $A=\left[\begin{array}{ccc}3 1 2 \\ 1 2 -3\end{array}\right]$ and $B=\left[\begin{array}{ccc}-2 0 4 \\ 5 -3 2\end{array}\right]$, find $(2 A-B)$ Solution: $2 A=2\left(\left[\begin{array}{ccc}3 1 2 \\ 1 2 -3\end{array}\right]\right)$ $=\left[\begin{array}{ccc}6 2 4 \\ 2 4 -6\end{array}\right]$ $(2 A-B)=\left[\begin{array}{ccc}6 2 4 \\ 2 4 -6\end{array}\right]-\left[\begin{array}{ccc}-2 0 4 \\ 5 -3 2\end{array}\right]$ $=\left[\begin{array}{ccc}8 2 0 \\ -3 7 -8\end{array}\right]$ Conc...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\sin x}{\sqrt{1+\sin x}} d x$ Solution: $\int \frac{\sin x}{\sqrt{1+\sin x}} d x$ We can write above integral as: $=\int \frac{1+\sin x-1}{\sqrt{1+\sin x}} d x$ (Adding and subtracting 1 in numerator) $=\int \frac{1+\sin x}{\sqrt{1+\sin x}} d x-\int \frac{1}{\sqrt{1+\sin x}} d x$ $=\int \sqrt{1+\sin x} d x-\int \frac{1}{\sqrt{1+\sin x}} d x$ Consider $\sqrt{1+\sin x}=\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}=\sqrt{\left(...
Read More →x = 3cosq – 2cos3q, y = 3sinq – 2sin3q.
Question: x =3cosq 2cos3q,y= 3sinq 2sin3q. Solution: Given,x =3cosq 2cos3q,y= 3sinq 2sin3q. Differentiating both the parametric functions w.r.t. q $\frac{d x}{d \theta}=-3 \sin \theta-6 \cos ^{2} \theta \cdot \frac{d}{d \theta}(\cos \theta)$ $=-3 \sin \theta-6 \cos ^{2} \theta \cdot(-\sin \theta)$ $=-3 \sin \theta+6 \cos ^{2} \theta \cdot \sin \theta$ $\frac{d y}{d \theta}=3 \cos \theta-6 \sin ^{2} \theta \cdot \frac{d}{d \theta}(\sin \theta)$ $=3 \cos \theta-6 \sin ^{2} \theta \cdot \cos \theta...
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Question: $x=e^{\theta}\left(\theta+\frac{1}{\theta}\right), y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$ Solution: Given, $x=e^{\theta}\left(\theta+\frac{1}{\theta}\right), y=e^{-\theta}\left(\theta-\frac{1}{\theta}\right)$ Differentiating both the parametric functions w.r.t. $\theta$. $\frac{d x}{d \theta}=e^{\theta}\left(1-\frac{1}{\theta^{2}}\right)+\left(\theta+\frac{1}{\theta}\right) \cdot e^{\theta}$ $\frac{d x}{d \theta}=e^{\theta}\left(1-\frac{1}{\theta^{2}}+\theta+\frac{1}{\theta...
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Question: $x=t+\frac{1}{t}, y=t-\frac{1}{t}$ Solution: Given, x = t + 1/t, y = t 1/t Differentiating both the parametric functions w.r.t $\frac{d x}{d t}=1-\frac{1}{t^{2}}, \frac{d y}{d t}=1+\frac{1}{t^{2}}$ $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1+\frac{1}{t^{2}}}{1-\frac{1}{t^{2}}}=\frac{t^{2}+1}{t^{2}-1}$ Thus, $\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1}$....
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Question: $\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right),-1x1, x \neq 0$ Solution: Let $y=\tan ^{-1}\left(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}\right)$ Putting $x^{2}=\cos 2 \theta \quad \therefore \theta=\frac{1}{2} \cos ^{-1} x^{2}$ $y=\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right)$ $y=\tan ^{-1}\left(\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{\cos (x-a) \cos (x-b)} d x$ Solution: Let $I=\int \frac{1}{\cos (x-a) \cos (x-b)} d x$ Multiply and divide $\frac{1}{\sin (a-b)}$ in R.H.S we get, $I=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\cos (x-a) \cos (x-b)} d x$ We can write above integral as: $=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b+x-x)}{\cos (x-a) \cos (x-b)} d x$ $=\frac{1}{\sin (a-b)} \int \frac{\sin [(x-b)-(x-a)]}{\cos (x-a) \cos (x-b)} d x$ $=\frac{1}{\sin (a-b)} \int\left[\frac{\sin (x-b) \c...
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Question: $\tan ^{-1}\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), \frac{-1}{\sqrt{3}}\frac{x}{a}\frac{1}{\sqrt{3}}$ Solution: Let $y=\tan ^{-1}\left[\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right]$ Put $x=a \tan \theta \quad \therefore \theta=\tan ^{-1} \frac{x}{a}$ $y=\tan ^{-1}\left[\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}\right]$ $y=\tan ^{-1}\left[\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{\sin (x-a) \sin (x-b)} d x$ Solution: Let $I=\int \frac{1}{\sin (x-a) \sin (x-b)} d x$ Multiply and divide $\frac{1}{\sin (a-b)}$ in R.H.S we get, $I=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\sin (x-a) \sin (x-b)} d x$ We can write above integral as: $=\frac{1}{\sin (a-b)} \int \frac{\sin (a-b+x-x)}{\sin (x-a) \sin (x-b)} d x$ $=\frac{1}{\sin (a-b)} \int \frac{\sin [(x-b)-(x-a)]}{\sin (x-a) \sin (x-b)} d x$ $=\frac{1}{\sin (a-b)} \int\left[\frac{\sin (x-b) \c...
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Question: $\sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right), 0x\frac{1}{\sqrt{2}}$ Solution: Let $y=\sec ^{-1}\left(\frac{1}{4 x^{3}-3 x}\right)$ Put $x=\cos \theta \quad \therefore \theta=\cos ^{-1} x$ $\Rightarrow y=\sec ^{-1}\left(\frac{1}{4 \cos ^{3} \theta-3 \cos \theta}\right)$ $y=\sec ^{-1}\left(\frac{1}{\cos 3 \theta}\right) \quad\left[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\right]$ $y=\sec ^{-1}(\sec 3 \theta) \Rightarrow y=3 \theta$ $y=3 \cos ^{-1} x$ Differentiating both s...
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Question: $\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right),-\frac{\pi}{2}x\frac{\pi}{2}$ and $\frac{a}{b} \tan x-1$ Solution: Let $y=\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right)$ $y=\tan ^{-1}\left[\frac{\frac{a \cos x}{b \cos x}-\frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x}+\frac{a \sin x}{b \cos x}}\right]$ $y=\tan ^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right]$ $y=\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x)$ $\left[\because \tan...
Read More →Evaluate the following integrals:
Question: $\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x$ Solution: $\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} d x$ We can write above integral as $=\int \frac{\sin x-\cos x}{\sqrt{1+\sin 2 x-1}} d x$ [Adding and subtracting 1 in denominator] $=\int \frac{\sin x-\cos x}{\sqrt{(1+\sin 2 x)-1}} d x$ $=\int \frac{\sin x-\cos x}{\sqrt{\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right)-1}} d x \because \sin ^{2} x+\cos ^{2} x=1$ and $\sin 2 x=2 \sin x \cos x$ $=\int \frac{(\sin x-\cos x)}{\sqrt{(...
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Question: $\tan ^{-1}(\sec x+\tan x),-\frac{\pi}{2}x\frac{\pi}{2}$ Solution: Let $y=\tan ^{-1}(\sec x+\tan x)$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x}\left[\tan ^{-1}(\sec x+\tan x)\right]$ $=\frac{1}{1+(\sec x+\tan x)^{2}} \cdot \frac{d}{d x}(\sec x+\tan x)$ $=\frac{1}{1+\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x} \cdot\left(\sec x \tan x+\sec ^{2} x\right)$ $=\frac{1}{\left(1+\tan ^{2} x\right)+\sec ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x)$ $=\frac{1}{\sec ...
Read More →Evaluate the following integrals:
Question: $\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$ Solution: $\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$ We can write above integral as $=\int \frac{\sin x+\cos x}{\sqrt{1-1+\sin 2 x}} d x$ [Adding and subtracting 1 in denominator] $=\int \frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2 x)}} d x$ $=\int \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}} d x \because \sin ^{2} x+\cos ^{2} x=1$ and $\sin 2 x=2 \sin x \cos x$ $=\int \frac{(\sin x+\cos x)}{\sqrt{1...
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Question: $\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right),-\frac{\pi}{4}x\frac{\pi}{4}$ Solution: Let $y=\tan ^{-1}\left[\sqrt{\frac{1-\cos x}{1+\cos x}}\right]$ $=\tan ^{-1}\left[\sqrt{\frac{2 \sin ^{2} x / 2}{2 \cos ^{2} x / 2}}\right]\left[\begin{array}{r}\because 1-\cos x=2 \sin ^{2} x / 2 \\ 1+\cos x=2 \cos ^{2} x / 2\end{array}\right]$ $=\tan ^{-1}\left[\frac{\sin x / 2}{\cos x / 2}\right]=\tan ^{-1}\left[\tan \frac{x}{2}\right]$ Thus, $y=\frac{x}{2}$ Differentiating both sides w....
Read More →Evaluate the following integrals:
Question: Evaluate $\int \cos x \cos 2 x \cos 3 x d x$ Solution: $\int \cos x \cos 2 x \cos 3 x d x$ We can write above integral as: $=\frac{1}{2} \int(2 \cos x \cos 2 x) \cos 3 x d x-(1)$ We know that, $2 \cos A \cdot \cos B=\cos (A+B)+\cos (A-B)$ Now, considering $A$ as $x$ and $B$ as $2 x$ we get, $=2 \cos x \cdot \cos 2 x=\cos (x+2 x)+\cos (x-2 x)$ $=2 \cos x \cdot \cos 2 x=\cos (3 x)+\cos (-x)$ $=2 \cos x \cdot \cos 2 x=\cos (3 x)+\cos (x)[\because \cos (-x)=\cos (x)]$ $\therefore$ integral...
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Question: $\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right), \frac{-\pi}{4}x\frac{\pi}{4}$ Solution: Let $y=\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)$ $=\cos ^{-1}\left[\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x\right]$ $=\cos ^{-1}\left[\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cdot \cos x\right]=\cos ^{-1}\left[\cos \left(\frac{\pi}{4}-x\right)\right]$ $y=\frac{\pi}{4}-x$ $\left[\because-\frac{\pi}{4}x\frac{\pi}{4}\right]$ Differentiating both sides w.r.t. $x$ Thu...
Read More →(x + 1)2 + (x + 2)3 + (x + 3)4
Question: (x+ 1)2+ (x+ 2)3+ (x+ 3)4 Solution: Let $\quad y=(x+1)^{2}(x+2)^{3}(x+3)^{4}$ Taking log on both sides, $\log y=\log \left[(x+1)^{2} \cdot(x+2)^{3} \cdot(x+3)^{4}\right]$ $\log y=\log (x+1)^{2}+\log (x+2)^{3}+\log (x+3)^{4}$ $\lceil\because \log x y=\log x+\log$ $\Rightarrow \log y=2 \log (x+1)+3 \log (x+2)+4 \log (x+3)$ $\left[\because \log x^{y}=y \log x\right]$ Differentiating both sides w.r.t. $x$; $\frac{1}{y} \cdot \frac{d y}{d x}=2 \cdot \frac{d}{d x} \log (x+1)+3 \cdot \frac{d}...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \sin x \sin 2 x \sin 3 x d x$ Solution: $\int \sin x \sin 2 x \sin 3 x d x$ We can write above integral as: $=\frac{1}{2} \int(2 \sin x \sin 2 x) \sin 3 x d x-(1)$ We know that, $2 \sin A \cdot \sin B=\cos (A-B)-\cos (A+B)$ Now, considering A as $x$ and B as $2 x$ we get, $=2 \sin x \cdot \sin 2 x=\cos (x-2 x)-\cos (x+2 x)$ $=2 \sin x \cdot \sin 2 x=\cos (-x)-\cos (3 x)$ $=2 \sin x \cdot \sin 2 x=\cos (x)-\cos (3 x)[\because \cos (-x)=\cos (x)]$ $\therefore$ integral (1)...
Read More →sinm x . cosn x
Question: sinmx . cosnx Solution: Let $y=\sin ^{m} x \cdot \cos ^{n} x$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x}\left(\sin ^{m} x \cdot \cos ^{n} x\right)$ $=\sin ^{m} x \cdot \frac{d}{d x}\left(\cos ^{n} x\right)+\cos ^{n} x \cdot \frac{d}{d x} \sin ^{m} x$ $=\sin ^{m} x \cdot n \cdot \cos ^{n-1} x \frac{d}{d x}(\cos x)+\cos ^{n} x \cdot m \cdot \sin ^{m-1} x$ $\frac{d}{d x}(\sin x)$ $=n \cdot \sin ^{m} x \cdot \cos ^{n-1} x \cdot(-\sin x)+m \cdot \cos ^{n} x \cdot \...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\cos ^{7} x}{\sin x} d x$ Solution: $\int \frac{\cos ^{7} x}{\sin x} d x$ We can write above integral as: $\int \frac{\left(\cos ^{2} x\right)^{3} \cdot \cos x}{\sin x} d x^{-(1)}$ Put $\operatorname{Sin} x=t$ Differentiting w.r.t $x$ we get, $\cos x \cdot d x=d t$ $\therefore$ integral $(1)$ becomes, $=\int \frac{\left(\cos ^{2} x\right)^{3}}{t} d t$ $=\int \frac{\left(1-\sin ^{2} x\right)^{2}}{t} d t \cdots\left(\because \sin ^{2}(x)+\cos ^{2}(x)=1\right)$ $=\int...
Read More →(sin x)cos x
Question: (sin x)cos x Solution: Let $y=(\sin x)^{\cos x}$ Taking log on both sides, $\log y=\log (\sin x)^{\cos x}$ $\Rightarrow \log y=\cos x \cdot \log (\sin x)$ $\left[\because \log x^{y}=y \log x\right]$ Differentiating both sides w.r.t. $x$, $\frac{1}{y} \cdot \frac{d y}{d x}=\frac{d}{d x} \cos x \cdot \log (\sin x)$ $\frac{1}{y} \cdot \frac{d y}{d x}=\cos x \cdot \frac{d}{d x} \log (\sin x)+\log (\sin x) \cdot \frac{d}{d x} \cos x$ $\frac{1}{y} \cdot \frac{d y}{d x}=\cos x \cdot \frac{1}{...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{e^{x}+e^{-x}} d x$ Solution: $\int \frac{1}{e^{x}+e^{-x}} d x$ We can write above integral as: $=\int \frac{1}{e^{x}+\frac{1}{e^{x}}} d x$ $=\int \frac{e^{x}}{e^{2 x}+1} d x-(1)$ Let $e^{x}=t$ Differentiating w.r.t $\mathrm{x}$ we get, $e^{x} d x=d t$ $\therefore$ integral (1) becomes, $=\tan ^{-1}(\mathrm{t})+C\left(\because \int \frac{1}{x^{2}+1} d x=\tan ^{-1}(x)\right)$ Putting value of $t$ we get, $=\tan ^{-1}\left(e^{x}\right)+C$ $\therefore \int \frac{1}{...
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