Which of the following is the principal
Question: Which of the following is the principal value branch of cos-1 x? (a) [-/2, /2] (b) (0, ) (c) [0. ] (d) [0, ] {/2} Solution: (c) [0. ] As we know that the principal value branch cos-1x is [0, ]....
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Question: Evaluate the following integral: $\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x$ Solution: Let, $I=\int \frac{x^{2}}{(x-1)\left(x^{2}+1\right)} d x$ Let $\frac{\mathrm{x}^{2}}{(\mathrm{x}-1)\left(\mathrm{x}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}-1}+\frac{\mathrm{B}}{\mathrm{x}^{2}+1}$ $\Rightarrow x^{2}=A\left(x^{2}+1\right)+B(x-1)$ For, $\mathrm{x}=1, \mathrm{~A}=\frac{1}{2}$ For, $\mathrm{x}=0, \mathrm{~B}=\frac{1}{2}$ $\therefore \mathrm{I}=\frac{1}{2} \int \frac{\mathrm{dx}}...
Read More →If a1, a2, a3, …., an is an arithmetic progression
Question: If a1, a2, a3, ., anis an arithmetic progression with common difference d, then evaluate the following expression. $\tan \left[\tan ^{-1}\left(\frac{d}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{d}{1+a_{2} a_{3}}\right)+\tan ^{-1}\left(\frac{d}{1+a_{3} a_{4}}\right)+\cdots+\tan ^{-1}\left(\frac{d}{1+a_{n-1} a_{n}}\right)\right]$ Solution: As a1, a2, a3, ., anis an arithmetic progression with common difference d. d = a2 a1= a3 a2= a4 a3= = an an-1 So, $\tan ^{-1} \frac{d}{1+a_{1} a_{2}...
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Question: Prove that: $2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$ Solution: To Prove: $2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31}=\frac{\pi}{4}$ Formula Used: 1) $2 \sin ^{-1} x=\sin ^{-1}\left(2 x x \sqrt{1-x^{2}}\right)$ 2) $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=2 \sin ^{-1} \frac{3}{5}-\tan ^{-1} \frac{17}{31} \ldots$ (1) $2 \sin ^{-1} \frac{3}{5}=\sin ^{-1}\left(2 \times \frac{3}{5} \times \sqrt{1-\l...
Read More →Show that tan-1 (1/2 sin-1 3/4) = (4 – √7)/ 3
Question: Show that tan-1(1/2 sin-13/4) = (4 7)/ 3 and justify why the other value (4 + 7)/ 3 is ignored. Solution: We have, tan-1(1/2 sin-13/4) Let sin-1 = ⇒ sin-1 = 2 ⇒ sin 2 = 2 tan / 1 + tan2 = 3 tan2 8 tan + 3 = 0 $\tan \theta=\frac{8 \pm \sqrt{64-36}}{6}$ $\tan \theta=\frac{8 \pm \sqrt{28}}{6}=\frac{8 \pm 2 \sqrt{7}}{6}=\frac{4 \pm \sqrt{7}}{3}$ Now, $-\frac{\pi}{2} \leq \sin ^{-1} \frac{3}{4} \leq \frac{\pi}{2}$ $\frac{-\pi}{4} \leq \frac{1}{2} \sin ^{-1} \frac{3}{4} \leq \frac{\pi}{2}$ $...
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Question: Evaluate the following integral: $\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+2\right)^{2}} d x$ Solution: Let, $\mathrm{I}=\int \frac{2 \mathrm{x}}{\left(\mathrm{x}^{2}+1\right)\left(\mathrm{x}^{2}+2\right)^{2}} \mathrm{dx}$ Let $x^{2}+2=t \Rightarrow 2 x d x=d t$ $\therefore \mathrm{I}=\int \frac{\mathrm{dt}}{(\mathrm{t}-1) \mathrm{t}^{2}}$ Now, let, $\frac{1}{(t-1) t^{2}}=\frac{A}{t-1}+\frac{B}{t}+\frac{C}{t^{2}}$ $\Rightarrow 1=A t^{2}+B t(t-1)+C(t-1)$ For $t=1, A=1$ For $t=0, ...
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Question: Find the value of 4 tan-11/5 tan-11/239 Solution: 4 tan-11/5 tan-11/239 = 2 (tan-11/5) tan-11/239 $=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$ $=2 \tan ^{-1} \frac{2 / 5}{24 / 25}-\tan ^{-1} \frac{1}{239}$ $=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}$ $=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because...
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Question: Evaluate the following integral: $\int \frac{1}{x^{4}-1} d x$ Solution: Let, $I=\int \frac{1}{\left(x^{4}-1\right)} d x$ Let $\frac{1}{\left(x^{4}-1\right)}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{x^{2}+1}$b $\Rightarrow 1=A(x-1)\left(x^{2}+1\right)+B(x+1)\left(x^{2}+1\right)+c(x+1)(x-1)$ For, $\mathrm{x}=1, \mathrm{~B}=\frac{1}{4}$ For, $x=-1, A=\frac{1}{4}$ For, $x=0, A=-\frac{1}{2}$ $\therefore \mathrm{I}=-\frac{1}{4} \int \frac{\mathrm{dx}}{\mathrm{x}+1}+\frac{1}{4} \int \frac{\mathrm...
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Question: Prove that: $\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{1}{2}$ Solution: To Prove: $\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{1}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}} \ldots$ (1) Let $\sin \theta=\frac{1}{\sqrt{17}}$ Therefore $\theta=\sin ^{-1} \frac{1}{\sqrt{17}} \ldots...
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Question: Evaluate the following integral: $\int \frac{1}{x\left(x^{4}-1\right)} d x$ Solution: Let, $\mathrm{I}=\int \frac{1}{\mathrm{x}\left(\mathrm{x}^{4}-1\right)} \mathrm{dx}$ Let, $\frac{1}{x\left(x^{4}-1\right)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}+\frac{D}{x^{2}+1}$ $\Rightarrow 1=A(x+1)(x-1)\left(x^{2}+1\right)+B x(x-1)\left(x^{2}+1\right)+c x(x+1)\left(x^{2}+1\right)+D x(x+1)(x-1)$ For, $x=0, A=-1$ For, $x=1, C=\frac{1}{4}$ $\therefore \mathrm{I}=-\int \frac{\mathrm{dx}}{\mathrm{x}}...
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Question: Prove that tan-11/4 + tan-12/9 = sin-11/5 Solution: Taking the LHS, tan-11/4 + tan-12/9 $=\tan ^{-1} \frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \cdot \frac{2}{9}}=\tan ^{-1} \frac{9+8}{36-2}=\tan ^{-1} \frac{1}{2}=\sin ^{-1} \frac{1}{\sqrt{5}}$ = RHS Hence Proved...
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Question: Show that sin-15/13 + cos-13/5 = tan-163/16 Solution: Here, sin-15/13 = tan-15/12 And, cos-13/5 = tan-14/3 Taking the L.H.S, we have L.H.S. $=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}$ $=\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{4}{3}=\tan ^{-1} \frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \cdot \frac{4}{3}}$ $=\tan ^{-1} \frac{\frac{15+48}{36}}{\frac{36-20}{36}}=\tan ^{-1} \frac{63}{16}$ Thus, L.H.S = R.H.S Hence Proved...
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Question: Prove that sin-18/17 + sin-13/5 = sin-177/85 Solution: Taking the L.H.S, = sin-18/17 + sin-13/5 $=\tan ^{-1} \frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}$ $=\tan ^{-1} \frac{\frac{32+45}{60}}{\frac{60-24}{60}}$ $=\tan ^{-1} 8 / 15+\tan ^{-1} 3 / 4$ $=\tan ^{-1} \frac{77}{36}$ $=\sin ^{-1} \frac{77}{\sqrt{5929+1296}}$ $=\sin ^{-1} \frac{77}{85}$ $=$ R.H.S. Hence proved...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$ Solution: Let, $I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+2\right)} d x$ Let, $x^{2}=y$ Then, $\frac{1}{(\mathrm{y}+1)(\mathrm{y}+2)}=\frac{\mathrm{A}}{\mathrm{y}+1}+\frac{\mathrm{B}}{\mathrm{y}+2}$ $\Rightarrow 1=A(y+2)+B(y+1)$ $\Rightarrow 1=(A+B) y+2 A+B$ On equating similar terms, we get, $A+B=0$, and $2 A+B=1$ We get, $A=1, B=-1$ $\therefore \mathrm{I}=\int \frac{\mathrm{dx}}{\math...
Read More →Find the simplified form
Question: Find the simplified form of cos-1[3/5 cos x + 4/5 sin x], x [-3/4, /4]. Solution: We have, cos-1[3/5 cos x + 4/5 sin x], x [-3/4, /4] Now, let cos = 3/5 So, sin = 4/5 and tan = 4/3 cos-1[3/5 cos x + 4/5 sin x] ⇒ cos-1[3/5 cos x + 4/5 sin x] = cos-1[cos cos x + sin sin x] = cos-1[cos ( x)] = x = tan-14/3 x...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{2 x+1}{(x-2)(x-3)} d x$ Solution: Let, $I=\int \frac{2 x+1}{(x-2)(x-3)} d x$ Now, let $\frac{2 x+1}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$ $\Rightarrow 2 x+1=A(x-3)+B(x-2)$ $\Rightarrow 2 x+1=(A+B) x-3 A-2 B$ Equating similar terms, we get, $A+B=2$ and $3 A+2 B=-1$ So, $A=-5, B=7$ $\therefore \mathrm{I}=-5 \int \frac{\mathrm{dx}}{\mathrm{x}-2}+7 \int \frac{\mathrm{dx}}{\mathrm{x}-3}$ $\Rightarrow 1=-5 \log |x-2|+7 \log |x-3|+c$ $\Rightarro...
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Question: Prove that $\tan ^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2}$ Solution: $\tan ^{-1} \frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2}$ Taking L.H.S, Let$x^{2}=\cos 2 \theta \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^{2}$ So, L.H.S. $=\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...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{\cos x}{(1-\sin x)^{3}(2+\sin x)} d x$ Solution: Let $\sin x=t$ $\cos x d x=d t$ $I=\int \frac{\cos x}{(1-\sin x)^{3}(2+\sin x)} d x$ $=\int \frac{d t}{(1-t)^{3}(2+t)}$ $\frac{1}{(1-t)^{3}(2+t)}=\frac{A}{1-t}+\frac{B}{(1-t)^{2}}+\frac{C}{(1-t)^{3}}+\frac{D}{2+t}$ $1=A(1-t)^{2}(2+t)+B(1-t)(2+t)+C(2+t)+D(1-t)^{3}$ Put $\mathrm{t}=1$ $1=3 C$ $C=\frac{1}{3}$ Put $t=-2$ $1=27 \mathrm{D}$ $D=\frac{1}{27}$ $A=-\frac{1}{27} B=\frac{1}{9}$ $\int \fra...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x$ Solution: $I=\int \frac{3}{(1-x)\left(1+x^{2}\right)} d x$ $\frac{3}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{B x+C}{1+x^{2}}$ $3=A\left(1+x^{2}\right)+(B x+C)(1-x)$ Equating similar terms $A-B=0$ $B-C=0$ $A+C=3$ Solving $A=\frac{3}{2}, B=\frac{3}{2}, C=\frac{3}{2}$ Thus, $I=\frac{3}{2} \int \frac{d x}{1-x}+\frac{3}{2} \int \frac{x d x}{1+x^{2}}+\frac{3}{2} \int \frac{d x}{1+x^{2}}$ $=-\frac{3}{2} \lo...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{x\left(x^{3}+8\right)} d x$ Solution: Consider the integral, $I=\int \frac{1}{x\left(x^{3}+8\right)} d x$ Rewriting the above integral, we have $I=\int \frac{x^{2}}{x^{3}\left(x^{3}+8\right)} d x$ $I=\frac{1}{3} \int \frac{3 x^{2}}{x^{3}\left(x^{3}+8\right)} d x$ Substitute $x^{3}=t$ $3 x^{2} d x=d t$ $I=\frac{1}{3} \int \frac{d t}{t(t+8)}$ $\frac{1}{t(t+8)}=\frac{A}{t}+\frac{B}{t+8}$ $1=A(t+8)+B t$ Equating constants $1=8 \mathrm{~A}$ $\...
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Question: Prove that: $\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{1}{2}$ Solution: To Prove: $\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}}=\tan ^{-1} \frac{1}{2}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=\sin ^{-1} \frac{1}{\sqrt{17}}+\cos ^{-1} \frac{9}{\sqrt{85}} \ldots$ (1) Let $\sin \theta=\frac{1}{\sqrt{17}}$ Therefore $\theta=\sin ^{-1} \frac{1}{\sqrt{1} 7} \ldot...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{x\left(x^{4}+1\right)} d x$ Solution: Let $I=\int \frac{1}{x\left(x^{4}+1\right)} d x$ $\frac{1}{x\left(x^{4}+1\right)}=\frac{A}{x}+\frac{B x^{3}+C x^{2}+D x+E}{x^{4}+1}$ $I=A\left(x^{4}+1\right)+\left(B x^{3}+C x^{2}+D x+E\right)(x)$ Equating constants $A=1$ Equating coefficients of $x^{4}$ $0=A+B$ $0=1+B$ $B=-1$ Equating coefficients of $x^{2}$ $D=0$ Equating coefficients of $x$ $E=0$ Thus, $I=\int \frac{d x}{x}+\int-\frac{x^{2} d x}{x^...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$ Solution: To Prove: $\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=\tan ^{-1} \frac{1}{3}+\sec ^{-1} \frac{\sqrt{5}}{2} \ldots$ (1) Let $\sec \theta=\frac{\sqrt{5}}{2}$ Therefore $\theta=\sec ^{-1} \frac{\sqrt{5}}{2} \cdots$ (2) From the figure, $\tan \theta=\frac{1}{2}...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$ Solution: $I=\int \frac{x^{2}+x+1}{(x+1)^{2}(x+2)} d x$ $\frac{\mathrm{x}^{2}+\mathrm{x}+1}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)}=\frac{\mathrm{A}}{\mathrm{x}+1}+\frac{\mathrm{B}}{(\mathrm{x}+1)^{2}}+\frac{\mathrm{c}}{\mathrm{x}+2}$ $x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}$ Put $x=-2$ $3=C$ $C=3$ Put $x=-1$ $1=B$ $B=1$ Equating coefficients of constants $1=2 A+2 B+C$ $1=2 A+2+3$ $A=-2$ Thus, $I=2 * \int \frac{\mat...
Read More →Solve the equation
Question: Solve the equation cos (tan-1x) = sin (cot-13/4). Solution: Given equation, cos (tan-1x) = sin (cot-13/4) Taking L.H.S, We have, $\cos \left(\tan ^{-1} x\right)=\sin \left(\cot ^{-1} \frac{3}{4}\right)$ L.H.S. $=\cos \left(\tan ^{-1} x\right)$ $=\cos \left(\cos ^{-1} \frac{1}{\sqrt{x^{2}+1}}\right)$ $=\frac{1}{\sqrt{x^{2}+1}}$ $\left(\because \cos \left(\cos ^{-1} x\right)=x, x \in[-1,1]\right)$ R.H.S. $=\sin \left(\cot ^{-1} \frac{3}{4}\right)$ $=\sin \left(\sin ^{-1} \frac{4}{5}\righ...
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