Prove the following
Question: $\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$ Solution: Let $y=\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^{2}}} \cdot \frac{d}{d x}\left(\frac{1}{\sqrt{x+1}}\right)$ $=\frac{1}{\sqrt{1-\frac{1}{x+1}}} \cdot \frac{d}{d x}(x+1)^{-1 / 2}$ $=\frac{1}{\sqrt{\frac{x+1-1}{x+1}}} \cdot \frac{-1}{2}(x+1)^{-3 / 2} \cdot \frac{d}{d x...
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Question: Evaluate $\int \frac{e^{x}-1}{e^{x}+1} d x$ Solution: $\int \frac{e^{x}-1}{e^{x}+1} d x$ We can write above integrand as: $\int\left(\frac{e^{x}}{e^{x}+1}-\frac{1}{e^{x}+1}\right) d x$ Considering integrand (A) $A=\int \frac{e^{x}}{e^{x}+1} d x$ Put $e^{x}+1=t$ Differentiating w.r.t $x$ we get, $e^{x} d x=d t$ Substituting values we get $A=\int \frac{e^{x}}{e^{x}+1} d x=\int \frac{d t}{t} d x=\ln |t|+C$ Substituting the value of t we get, $A=\ln \left|e^{x}+1\right|+C$ $\therefore A=\i...
Read More →sin x2 + sin2 x + sin2 (x2)
Question: sin x2+ sin2x + sin2(x2) Solution: Let $y=\sin x^{2}+\sin ^{2} x+\sin ^{2}\left(x^{2}\right)$ Differentiating both sides w.r.t. $x$, $\frac{d y}{d x}=\frac{d}{d x} \sin \left(x^{2}\right)+\frac{d}{d x} \sin ^{2} x+\frac{d}{d x} \sin ^{2}\left(x^{2}\right)$ $=\cos x^{2} \cdot \frac{d}{d x}\left(x^{2}\right)+2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \sin \left(x^{2}\right) \frac{d}{d x} \sin \left(x^{2}\right)$ $=\cos x^{2} \cdot 2 x+2 \sin x \cdot \cos x+2 \sin x^{2} \cdot \cos x^{2} \cdot...
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Question: $\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$ Solution: Let $y=\sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}\left(\frac{1}{\sqrt{x+1}}\right)=\frac{1}{\sqrt{1-\left(\frac{1}{\sqrt{x+1}}\right)^{2}}} \cdot \frac{d}{d x}\left(\frac{1}{\sqrt{x+1}}\right)$ $=\frac{1}{\sqrt{1-\frac{1}{x+1}}} \cdot \frac{d}{d x}(x+1)^{-1 / 2}$ $=\frac{1}{\sqrt{\frac{x+1-1}{x+1}}} \cdot \frac{-1}{2}(x+1)^{-3 / 2} \cdot \frac{d}{d x...
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Question: Evaluate $\int \frac{1}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx}$ Solution: $\int \frac{1}{e^{x}+1} d x$ We can write above integral as $\Rightarrow \int \frac{1+e^{x}-e^{x}}{e^{x}+1} d x$ Considering first integral: $\int \frac{1+e^{x}}{1+e^{x}} d x$ Since the numerator and denominator are exactly same, our integrand simplifies to 1 and integrand becomes: $\Rightarrow \int d x$ $\Rightarrow x$ $\therefore \int \frac{1+e^{x}}{1+e^{x}} d x=x \cdots$ (3) Considering second integral: $\int ...
Read More →sin x2 + sin2 x + sin2 (x2)
Question: sin x2+ sin2x + sin2(x2) Solution: Let $y=\sin x^{2}+\sin ^{2} x+\sin ^{2}\left(x^{2}\right)$ Differentiating both sides w.r.t. $x$, $\frac{d y}{d x}=\frac{d}{d x} \sin \left(x^{2}\right)+\frac{d}{d x} \sin ^{2} x+\frac{d}{d x} \sin ^{2}\left(x^{2}\right)$ $=\cos x^{2} \cdot \frac{d}{d x}\left(x^{2}\right)+2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \sin \left(x^{2}\right) \frac{d}{d x} \sin \left(x^{2}\right)$ $=\cos x^{2} \cdot 2 x+2 \sin x \cdot \cos x+2 \sin x^{2} \cdot \cos x^{2} \cdot...
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Question: $\cos (\tan \sqrt{x+1})$ Solution: Let $y=\cos (\tan \sqrt{x+1})$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x} \cos (\tan \sqrt{x+1})$ $=-\sin (\tan \sqrt{x+1}) \cdot \frac{d}{d x}(\tan \sqrt{x+1})$ $=-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \cdot \frac{d}{d x} \sqrt{x+1}$ $=-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \cdot \frac{1}{2 \sqrt{x+1}} \cdot 1$ Thus, $\quad \frac{d y}{d x}=-\frac{1}{2 \sqrt{x+1}} \sin (\tan \sqrt{x+1}) \cdot \sec ^{2}...
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Question: Evaluate $\int \frac{\left\{\cot x+\cot ^{3}\right\} x}{1+\cot ^{3} x} d x$ Solution: $=\int \frac{\cot x\left(1+\cot ^{2} x\right)}{1+\cot ^{3} x} d x=\int \frac{\cot x \operatorname{cosec}^{2} x}{1+\cot ^{3} x} d x$ Put $\cot x=t,-\operatorname{cosec}^{2} x d x=d t$ $=-\int \frac{t d t}{t^{3}+1}=-\int \frac{t d t}{(t+1)\left(t^{2}-t+1\right)}$ By partial fractions it's a remembering thing That if you see the above integral just apply the below return result, $=-\int\left[\frac{(t+1)}...
Read More →sinn (ax2 + bx + c)
Question: sinn(ax2+ bx + c) Solution: Let $y=\sin ^{n}\left(a x^{2}+b x+c\right)$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x} \sin ^{n}\left(a x^{2}+b x+c\right)$ $=n \cdot \sin ^{n-1}\left(a x^{2}+b x+c\right) \cdot \frac{a}{d x} \sin \left(a x^{2}+b x+c\right)$ $=n \cdot \sin ^{n-1}\left(a x^{2}+b x+c\right) \cdot \cos \left(a x^{2}+b x+c\right) \cdot \frac{d}{d x}\left(a x^{2}+b x+c\right)$ $=n \cdot \sin ^{n-1}\left(a x^{2}+b x+c\right) \cdot \cos \left(a x^{2}+b x+c...
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Question: $\int \frac{\sin 4 x-2}{1-\cos 4 x} e^{2 x} d x$ Solution: Put $2 x=t$; $2 d x=d t ; d x=d t / 2$ $=-\int \frac{\sin 4 x-2}{\cos 4 x-1} d x=-\frac{1}{2} \int \frac{e^{t}(\sin 2 t-2)}{\cos 2 t-1} d t=\frac{1}{4} \int \frac{e^{t}(2 \sin t \cos t-2)}{\cos ^{2} t} d t$ $=\frac{2}{4} \int e^{t} \cot t d t-\frac{2}{4} \int e^{t} \operatorname{cosec}^{2} t d t=\frac{1}{2}\left[\int e^{t} \operatorname{cottdt}-\int e^{t} \operatorname{cosec}^{2} t d t\right]$ $=\frac{1}{2}\left[e^{t} \cot t+\i...
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Question: $\sin \sqrt{x}+\cos ^{2} \sqrt{x}$ Solution: Let $y=\sin \sqrt{x}+\cos ^{2} \sqrt{x}$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x}(\sin \sqrt{x})+\frac{d}{d x}\left(\cos ^{2} \sqrt{x}\right)$ $=\cos \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x})+2 \cos \sqrt{x} \cdot \frac{d}{d x}(\cos \sqrt{x})$ $=\cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}+2 \cos \sqrt{x}(-\sin \sqrt{x}) \cdot \frac{d}{d x} \sqrt{x}$ $=\frac{1}{2 \sqrt{x}} \cdot \cos \sqrt{x}-2 \cos \sqrt{x} \cdot \sin \...
Read More →Evaluate the following integrals:
Question: $\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$ Solution: $=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$ by partial fraction, $\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2}$ So we get these three equations, $2 \mathrm{~A}+2 \mathrm{~B}+\mathrm{C}=1$ $3 \mathrm{~A}+\mathrm{B}+2 \mathrm{C}=1$ $\mathrm{A}+\mathrm{C}=1$ So the values are $A=-2 ; C=3 ; B=1$ $\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x=\int\left(-\frac{2 d x}{x+1}\right)+\int \frac{d x}{(x+1)^{2}}+...
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Question: Evaluate $\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x$ Solution: Let, $x=\sin ^{2} t$ Differentiating both side with respect to t $\frac{d x}{d t}=2 \sin t \cos t \Rightarrow d x=2 \sin t \cos t d t$ $y=\int \sqrt{\frac{1-\sin t}{1+\sin t}} 2 \sin t \cos t d t$ $y=\int \sqrt{\frac{(1-\sin t)}{(1+\sin t)} \times \frac{(1-\sin t)}{(1-\sin t)}} 2 \sin t \cos t d t$ $y=2 \int(1-\sin t) \sin t d t$ $y=2 \int \sin t-\frac{1-\cos 2 t}{2} d t$ $y=2\left(-\cos t-\frac{t}{2}+\frac{\sin 2 t}{4}...
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Question: $\int \frac{x^{2}-2}{x^{5}-x} d x$ Solution: By partial fractions, $=\frac{x^{2}-2}{x^{2}-5}=\frac{x^{2}-2}{(x-1) x(x+1)\left(x^{2}+1\right)}=\frac{A}{x-1}+\frac{B}{x}+\frac{C}{x+1}+\frac{D}{x^{2}+1}$ $=\int-\frac{d x}{4(x-1)}+\int \frac{2}{x} d x-\frac{1}{4} \int \frac{d x}{x+1}-\frac{3}{2} \int \frac{x d x}{x^{2}+1}$ $=-\frac{1}{4} \log (x-1)+2 \log x-\frac{1}{4} \log (x+1)-\frac{3}{4} \log \left(x^{2}+1\right)+c$...
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Question: $\log \left[\log \left(\log x^{5}\right)\right]$ Solution: Let, $\quad y=\log \left[\log \left(\log x^{5}\right)\right]$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x} \log \left[\log \left(\log x^{5}\right)\right]$ $=\frac{1}{\log \left(\log x^{5}\right)} \times \frac{d}{d x} \log \left(\log x^{5}\right)$ $=\frac{1}{\log \left(\log x^{5}\right)} \times \frac{1}{\log \left(x^{5}\right)} \times \frac{d}{d x} \log x^{5}$ $=\frac{1}{\log \left(\log x^{5}\right)} \cdo...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{\left(x^{2}+2\right)\left(x^{2}+5\right)} d x$ Solution: $\int \frac{d x}{\left(x^{2}+5\right)\left(x^{2}+2\right)}$ By partial fractions, $\frac{1}{\left(x^{2}+5\right)\left(x^{2}+2\right)}=\frac{A}{x^{2}+5}+\frac{B}{x^{2}+2}$ Solving these two equations, $2 A+5 B=1$ and $A+B=0$ We get $A=-1 / 3$ and $B=1 / 3$ $=-\frac{1}{3} \int \frac{d x}{\left(x^{2}+5\right)}+\frac{1}{3} \int \frac{d x}{\left(x^{2}+2\right)}=-\frac{1}{3} \cdot \frac{1}{\sqrt{5}} \tan ^{-1} \...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{1}{1+x+x^{2}+x^{3}} d x$ Solution: $=\int \frac{d x}{1+x+x^{2}+x^{3}}=\int \frac{d x}{(1+x)\left(1+x^{2}\right)}$ We can write the integral as follows, $=\int\left[\frac{d x}{2(x+1)}\right]-\int\left[\frac{x-1}{2\left(x^{2}+1\right)} d x\right]=\frac{1}{2} \log (x+1)-\frac{1}{2}\left[\int \frac{x d x}{x^{2}+1}-\int \frac{d x}{x^{2}+1}\right]$ $=\frac{1}{2} \log (x+1)-\frac{1}{2}\left[\log \frac{\left(x^{2}+1\right)}{2}-\tan ^{-1} x\right]+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{x}{x^{3}-1} d x$ Solution: $=\int \frac{x}{\left(x^{3}-1\right)} d x=\int \frac{x}{(x-1)\left(x^{2}+x+1\right)} d x$ $=\int\left(\frac{1}{3(x-1)}-\frac{x-1}{3\left(x^{2}+x+1\right)}\right)$ $=\frac{1}{3} \int \frac{1}{x-1} d x-\frac{1}{3} \int \frac{x-1}{x^{2}+x+1} d x$ $=\frac{1}{3} \log (x-1)-\frac{1}{3}\left[\int \frac{(2 x+1)}{2\left(x^{2}+x+1\right)} d x-\int \frac{3}{2\left(\left(x^{2}+x+1\right)\right)} d x\right]$' $=\frac{1}{3} \log (x-1)-\frac{1}{3}[I 1+I...
Read More →Prove the following
Question: $\log \left(x+\sqrt{x^{2}+a}\right)$ Solution: Let $y=\log \left(x+\sqrt{x^{2}+a}\right)$ Differentiating both sides w.r.t. $x$ $\frac{d y}{d x}=\frac{d}{d x} \log \left(x+\sqrt{x^{2}+a}\right)$ $=\frac{1}{x+\sqrt{x^{2}+a}} \cdot \frac{d}{d x}\left(x+\sqrt{x^{2}+a}\right)$ $=\frac{1}{x+\sqrt{x^{2}+a}} \cdot\left[1+\frac{1}{2 \sqrt{x^{2}+a}} \times \frac{d}{d x}\left(x^{2}+a\right)\right]$ $=\frac{1}{x+\sqrt{x^{2}+a}} \cdot\left[1+\frac{1}{2 \sqrt{x^{2}+a}} \cdot 2 x\right]$ $=\frac{1}{...
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Question: $\frac{8^{x}}{x^{8}}$ Solution: Let $y=\frac{8^{x}}{x^{8}}$ Taking log on both sides, we get $\log y=\log \frac{8^{x}}{x^{8}}$ $\Rightarrow \log y=\log 8^{x}-\log x^{8} \Rightarrow \log y=x \log 8-8 \log x$ Differentiating both sides w.r.t. $x$ $\frac{1}{y} \cdot \frac{d y}{d x}=\log 8.1-\frac{8}{x} \Rightarrow \frac{d y}{d x}=y\left[\log 8-\frac{8}{x}\right]$ Thus, $\frac{d y}{d x}=\frac{8^{x}}{x^{8}}\left[\log 8-\frac{8}{x}\right]$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$ Solution: $=\int \frac{x^{2}}{(x-1)^{3}(x+1)} d x$\ By using partial differentiation, $=\frac{x^{2}}{(x-1)^{3}(x+1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}+\frac{C}{(x-1)^{2}}+\frac{D}{(x-1)^{3}}$ $x^{2}=A(x-1)^{3}+B(x-1)^{2}(x+1)+C(x-1)^{1}(x+1)+D(x+1)$ By substituting the $x^{2}$ coefficients and other coefficients we can get, $\mathrm{A}=-1 / 8 ; \mathrm{B}=1 / 8 ; \mathrm{C}=3 / 4 ; \mathrm{D}=1 / 2$ $=\int \frac{-d x}{8(x+1)}+\int \frac{d ...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\mathrm{e}^{\mathrm{m} \tan ^{-1} \mathrm{x}}}{\left(1+\mathrm{x}^{2}\right)^{3 / 2}} \mathrm{dx}$ Solution: $=e^{m} \int \frac{\tan ^{-1} x}{\left(1+x^{2}\right) \sqrt{1+x^{2}}} d x$ Put $\tan ^{-1} x=t, d x /\left(1+x^{2}\right)=d t, 1+x^{2}=\sec ^{2} x$; $=e^{m} \int \frac{t d t}{\sec t}=e^{m} \int t \cos t d t$ $=e^{m}\left[t \sin t-\int \sin t d t\right]$ $=e^{m}[t \sin t+\cos t]+c$ $=e^{m}\left[\frac{x \tan ^{-1} x}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+x^{2}}}\ri...
Read More →Prove the following
Question: $2^{\cos ^{2} x}$ Solution: Let $y=2^{\cos ^{2} x}$ Taking log on both sides, we get $\log y=\log 2^{\cos ^{2} x} \Rightarrow \log y=\cos ^{2} x \cdot \log 2$ Now, Differentiating both sides w.r.t. $x$ $\frac{1}{y} \cdot \frac{d y}{d x}=\log 2 \cdot \frac{d}{d x} \cos ^{2} x$ $\frac{1}{y} \cdot \frac{d y}{d x}=\log 2\left[2 \cos x \cdot \frac{d}{d x} \cos x\right]$ $\frac{1}{y} \cdot \frac{d y}{d x}=\log 2[2 \cos x(-\sin x)]$ $\frac{1}{y} \cdot \frac{d y}{d x}=\log 2(-\sin 2 x)$ $\frac...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \mathrm{e}^{\mathrm{x}} \frac{(1-\mathrm{x})^{2}}{\left(1+\mathrm{x}^{2}\right)^{2}} \mathrm{dx}$ Solution: $=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}}$ $=\int e^{x} \frac{d x}{1+x^{2}}-\int \frac{2 x e^{x} d x}{\left(1+x^{2}\right)^{2}}$ $=\int e^{x}\left[\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x \ldots \ldots\left(\int e^{x}\left(f(x)+f^{\prime}(x)\right)=e^{x} f(x)+c\right)$ $=e^{x} \frac{1}{1+x^{2}}+c$...
Read More →Evaluate the following integrals:
Question: Evaluate $\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{-x / 2} d x$ Solution: $=\int e^{\frac{x}{2}} \frac{\sqrt{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}}{2 \cos ^{2} \frac{x}{2}}=$ $\int e^{-\frac{x}{2}} \frac{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)}{2 \cos ^{2} \frac{x}{2}} d x$ $=\int e^{-\frac{x}{2}}\left(\frac{\sin \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}-\frac{\cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right) d x$ $=\int\left[\frac{1}{2} \t...
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