86325
If \(\int \mathrm{e}^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right)\) \(d x=e^{\sec x} f(x)+C\), then a possible choice of \(f(x)\) is
1 \(x \sec x+\tan x+\frac{1}{2}\)
2 \(\sec x+\tan x+\frac{1}{2}\)
3 \(\sec x+x \tan x-\frac{1}{2}\)
4 \(\sec x-\tan x-\frac{1}{2}\)
Explanation:
(B) : Given, \(\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x=e^{\sec x} f(x)+C\) On differentiating both side w. r. t. \(x\), we get - \(\left.\mathrm{e}^{\sec x}(\sec x \tan x f(x))+\sec x \tan x \sec ^{2} x\right)\) \(=e^{\sec x} \sec x \tan x f(x)+e^{\sec x} f^{\prime}(x)\) \(\sec x \tan x+\sec ^{2} x=f^{\prime}(x)\) \(f(x)=\sec x+\tan x+\frac{1}{2}\)
JEE Main-2019-09.04.2019
Integral Calculus
86326
If \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\mathbf{A}\left(\tan ^{-1}\left(\frac{\mathbf{x}-1}{3}\right)+\frac{\mathbf{f}(\mathbf{x})}{\mathbf{x}^{2}-\mathbf{2 x}+\mathbf{1 0}}\right)+\mathbf{C}\) where, \(C\) is a constant of integration, then
1 \(\mathrm{A}=\frac{1}{27}\) and \(\mathrm{f}(\mathrm{x})=9(\mathrm{x}-1)\)
2 \(A=\frac{1}{81}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
3 \(\mathrm{A}=\frac{1}{54}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
4 \(A=\frac{1}{54}\) and \(f(x)=9(x-1)^{2}\)
Explanation:
(C) : Given, \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=A\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+C\) L.H.S \(I=\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\int \frac{\mathrm{dx}}{\left((\mathrm{x}-1)^{2}+9\right)^{2}}\) Let, \(x-1=3 \tan t\) \(\mathrm{dx}=3 \sec ^{2} \mathrm{t} d \mathrm{t}\) \(I=\int \frac{3 \sec ^{2} t}{\left(9 \tan ^{2}+9\right)^{2}} d t\) \(=\int \frac{3 \sec ^{2} t}{81 \sec ^{4} t} d t=\int \frac{\cos ^{2} t}{27} d t\) \(=\int \frac{1+\cos 2 t}{27} d t=\frac{1}{54}\left(t+\frac{\sin 2 t}{2}\right)+C\) \(=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x-3}{3}\right)+\frac{3(x-1)}{x^{2}-2 x+10}\right]+C\) \(\therefore \quad A =\frac{1}{54} \text { and } f(x)=3(x-1)\)
JEE Main-2019-10.04.2019
Integral Calculus
86334
\(\int \sin (11 x) \cdot \sin ^{9} x d x=\) \(+\mathbf{C}\).
(A) : \(\mathrm{I}=\int \sin (11 \mathrm{x}) \cdot \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int \sin (10 x+x) \cdot \sin ^{9} x d x\) \(I=\int(\sin 10 x \cos x+\cos 10 x \sin x) \sin ^{9} x d x\) \(I=\int \sin 10 x \cos x \cdot \sin ^{9} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) Integration by parts - (From I LATE) \(I=\sin 10 x \cdot \int \sin ^{9} x \cos x d x-\int \frac{d}{d x} \sin 10 x \int \sin ^{9} x \cos x d x\) \(+\int \cos 10 x \cdot \sin ^{10} x d x\) \(=\sin 10 x \cdot \frac{\sin ^{10} x}{10}-\frac{10}{10} \int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x=\) \(=\frac{\sin 10 x \cdot \sin ^{10} x}{10}-\int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) \(I=\frac{\sin 10 x \cdot \sin ^{10} x}{10}+C\)
86325
If \(\int \mathrm{e}^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right)\) \(d x=e^{\sec x} f(x)+C\), then a possible choice of \(f(x)\) is
1 \(x \sec x+\tan x+\frac{1}{2}\)
2 \(\sec x+\tan x+\frac{1}{2}\)
3 \(\sec x+x \tan x-\frac{1}{2}\)
4 \(\sec x-\tan x-\frac{1}{2}\)
Explanation:
(B) : Given, \(\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x=e^{\sec x} f(x)+C\) On differentiating both side w. r. t. \(x\), we get - \(\left.\mathrm{e}^{\sec x}(\sec x \tan x f(x))+\sec x \tan x \sec ^{2} x\right)\) \(=e^{\sec x} \sec x \tan x f(x)+e^{\sec x} f^{\prime}(x)\) \(\sec x \tan x+\sec ^{2} x=f^{\prime}(x)\) \(f(x)=\sec x+\tan x+\frac{1}{2}\)
JEE Main-2019-09.04.2019
Integral Calculus
86326
If \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\mathbf{A}\left(\tan ^{-1}\left(\frac{\mathbf{x}-1}{3}\right)+\frac{\mathbf{f}(\mathbf{x})}{\mathbf{x}^{2}-\mathbf{2 x}+\mathbf{1 0}}\right)+\mathbf{C}\) where, \(C\) is a constant of integration, then
1 \(\mathrm{A}=\frac{1}{27}\) and \(\mathrm{f}(\mathrm{x})=9(\mathrm{x}-1)\)
2 \(A=\frac{1}{81}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
3 \(\mathrm{A}=\frac{1}{54}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
4 \(A=\frac{1}{54}\) and \(f(x)=9(x-1)^{2}\)
Explanation:
(C) : Given, \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=A\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+C\) L.H.S \(I=\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\int \frac{\mathrm{dx}}{\left((\mathrm{x}-1)^{2}+9\right)^{2}}\) Let, \(x-1=3 \tan t\) \(\mathrm{dx}=3 \sec ^{2} \mathrm{t} d \mathrm{t}\) \(I=\int \frac{3 \sec ^{2} t}{\left(9 \tan ^{2}+9\right)^{2}} d t\) \(=\int \frac{3 \sec ^{2} t}{81 \sec ^{4} t} d t=\int \frac{\cos ^{2} t}{27} d t\) \(=\int \frac{1+\cos 2 t}{27} d t=\frac{1}{54}\left(t+\frac{\sin 2 t}{2}\right)+C\) \(=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x-3}{3}\right)+\frac{3(x-1)}{x^{2}-2 x+10}\right]+C\) \(\therefore \quad A =\frac{1}{54} \text { and } f(x)=3(x-1)\)
JEE Main-2019-10.04.2019
Integral Calculus
86334
\(\int \sin (11 x) \cdot \sin ^{9} x d x=\) \(+\mathbf{C}\).
(A) : \(\mathrm{I}=\int \sin (11 \mathrm{x}) \cdot \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int \sin (10 x+x) \cdot \sin ^{9} x d x\) \(I=\int(\sin 10 x \cos x+\cos 10 x \sin x) \sin ^{9} x d x\) \(I=\int \sin 10 x \cos x \cdot \sin ^{9} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) Integration by parts - (From I LATE) \(I=\sin 10 x \cdot \int \sin ^{9} x \cos x d x-\int \frac{d}{d x} \sin 10 x \int \sin ^{9} x \cos x d x\) \(+\int \cos 10 x \cdot \sin ^{10} x d x\) \(=\sin 10 x \cdot \frac{\sin ^{10} x}{10}-\frac{10}{10} \int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x=\) \(=\frac{\sin 10 x \cdot \sin ^{10} x}{10}-\int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) \(I=\frac{\sin 10 x \cdot \sin ^{10} x}{10}+C\)
86325
If \(\int \mathrm{e}^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right)\) \(d x=e^{\sec x} f(x)+C\), then a possible choice of \(f(x)\) is
1 \(x \sec x+\tan x+\frac{1}{2}\)
2 \(\sec x+\tan x+\frac{1}{2}\)
3 \(\sec x+x \tan x-\frac{1}{2}\)
4 \(\sec x-\tan x-\frac{1}{2}\)
Explanation:
(B) : Given, \(\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x=e^{\sec x} f(x)+C\) On differentiating both side w. r. t. \(x\), we get - \(\left.\mathrm{e}^{\sec x}(\sec x \tan x f(x))+\sec x \tan x \sec ^{2} x\right)\) \(=e^{\sec x} \sec x \tan x f(x)+e^{\sec x} f^{\prime}(x)\) \(\sec x \tan x+\sec ^{2} x=f^{\prime}(x)\) \(f(x)=\sec x+\tan x+\frac{1}{2}\)
JEE Main-2019-09.04.2019
Integral Calculus
86326
If \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\mathbf{A}\left(\tan ^{-1}\left(\frac{\mathbf{x}-1}{3}\right)+\frac{\mathbf{f}(\mathbf{x})}{\mathbf{x}^{2}-\mathbf{2 x}+\mathbf{1 0}}\right)+\mathbf{C}\) where, \(C\) is a constant of integration, then
1 \(\mathrm{A}=\frac{1}{27}\) and \(\mathrm{f}(\mathrm{x})=9(\mathrm{x}-1)\)
2 \(A=\frac{1}{81}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
3 \(\mathrm{A}=\frac{1}{54}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
4 \(A=\frac{1}{54}\) and \(f(x)=9(x-1)^{2}\)
Explanation:
(C) : Given, \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=A\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+C\) L.H.S \(I=\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\int \frac{\mathrm{dx}}{\left((\mathrm{x}-1)^{2}+9\right)^{2}}\) Let, \(x-1=3 \tan t\) \(\mathrm{dx}=3 \sec ^{2} \mathrm{t} d \mathrm{t}\) \(I=\int \frac{3 \sec ^{2} t}{\left(9 \tan ^{2}+9\right)^{2}} d t\) \(=\int \frac{3 \sec ^{2} t}{81 \sec ^{4} t} d t=\int \frac{\cos ^{2} t}{27} d t\) \(=\int \frac{1+\cos 2 t}{27} d t=\frac{1}{54}\left(t+\frac{\sin 2 t}{2}\right)+C\) \(=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x-3}{3}\right)+\frac{3(x-1)}{x^{2}-2 x+10}\right]+C\) \(\therefore \quad A =\frac{1}{54} \text { and } f(x)=3(x-1)\)
JEE Main-2019-10.04.2019
Integral Calculus
86334
\(\int \sin (11 x) \cdot \sin ^{9} x d x=\) \(+\mathbf{C}\).
(A) : \(\mathrm{I}=\int \sin (11 \mathrm{x}) \cdot \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int \sin (10 x+x) \cdot \sin ^{9} x d x\) \(I=\int(\sin 10 x \cos x+\cos 10 x \sin x) \sin ^{9} x d x\) \(I=\int \sin 10 x \cos x \cdot \sin ^{9} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) Integration by parts - (From I LATE) \(I=\sin 10 x \cdot \int \sin ^{9} x \cos x d x-\int \frac{d}{d x} \sin 10 x \int \sin ^{9} x \cos x d x\) \(+\int \cos 10 x \cdot \sin ^{10} x d x\) \(=\sin 10 x \cdot \frac{\sin ^{10} x}{10}-\frac{10}{10} \int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x=\) \(=\frac{\sin 10 x \cdot \sin ^{10} x}{10}-\int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) \(I=\frac{\sin 10 x \cdot \sin ^{10} x}{10}+C\)
86325
If \(\int \mathrm{e}^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right)\) \(d x=e^{\sec x} f(x)+C\), then a possible choice of \(f(x)\) is
1 \(x \sec x+\tan x+\frac{1}{2}\)
2 \(\sec x+\tan x+\frac{1}{2}\)
3 \(\sec x+x \tan x-\frac{1}{2}\)
4 \(\sec x-\tan x-\frac{1}{2}\)
Explanation:
(B) : Given, \(\int e^{\sec x}\left(\sec x \tan x f(x)+\left(\sec x \tan x+\sec ^{2} x\right)\right) d x=e^{\sec x} f(x)+C\) On differentiating both side w. r. t. \(x\), we get - \(\left.\mathrm{e}^{\sec x}(\sec x \tan x f(x))+\sec x \tan x \sec ^{2} x\right)\) \(=e^{\sec x} \sec x \tan x f(x)+e^{\sec x} f^{\prime}(x)\) \(\sec x \tan x+\sec ^{2} x=f^{\prime}(x)\) \(f(x)=\sec x+\tan x+\frac{1}{2}\)
JEE Main-2019-09.04.2019
Integral Calculus
86326
If \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\mathbf{A}\left(\tan ^{-1}\left(\frac{\mathbf{x}-1}{3}\right)+\frac{\mathbf{f}(\mathbf{x})}{\mathbf{x}^{2}-\mathbf{2 x}+\mathbf{1 0}}\right)+\mathbf{C}\) where, \(C\) is a constant of integration, then
1 \(\mathrm{A}=\frac{1}{27}\) and \(\mathrm{f}(\mathrm{x})=9(\mathrm{x}-1)\)
2 \(A=\frac{1}{81}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
3 \(\mathrm{A}=\frac{1}{54}\) and \(\mathrm{f}(\mathrm{x})=3(\mathrm{x}-1)\)
4 \(A=\frac{1}{54}\) and \(f(x)=9(x-1)^{2}\)
Explanation:
(C) : Given, \(\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}=A\left(\tan ^{-1}\left(\frac{x-1}{3}\right)+\frac{f(x)}{x^{2}-2 x+10}\right)+C\) L.H.S \(I=\int \frac{d x}{\left(x^{2}-2 x+10\right)^{2}}\) \(=\int \frac{\mathrm{dx}}{\left((\mathrm{x}-1)^{2}+9\right)^{2}}\) Let, \(x-1=3 \tan t\) \(\mathrm{dx}=3 \sec ^{2} \mathrm{t} d \mathrm{t}\) \(I=\int \frac{3 \sec ^{2} t}{\left(9 \tan ^{2}+9\right)^{2}} d t\) \(=\int \frac{3 \sec ^{2} t}{81 \sec ^{4} t} d t=\int \frac{\cos ^{2} t}{27} d t\) \(=\int \frac{1+\cos 2 t}{27} d t=\frac{1}{54}\left(t+\frac{\sin 2 t}{2}\right)+C\) \(=\frac{1}{54}\left[\tan ^{-1}\left(\frac{x-3}{3}\right)+\frac{3(x-1)}{x^{2}-2 x+10}\right]+C\) \(\therefore \quad A =\frac{1}{54} \text { and } f(x)=3(x-1)\)
JEE Main-2019-10.04.2019
Integral Calculus
86334
\(\int \sin (11 x) \cdot \sin ^{9} x d x=\) \(+\mathbf{C}\).
(A) : \(\mathrm{I}=\int \sin (11 \mathrm{x}) \cdot \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int \sin (10 x+x) \cdot \sin ^{9} x d x\) \(I=\int(\sin 10 x \cos x+\cos 10 x \sin x) \sin ^{9} x d x\) \(I=\int \sin 10 x \cos x \cdot \sin ^{9} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) Integration by parts - (From I LATE) \(I=\sin 10 x \cdot \int \sin ^{9} x \cos x d x-\int \frac{d}{d x} \sin 10 x \int \sin ^{9} x \cos x d x\) \(+\int \cos 10 x \cdot \sin ^{10} x d x\) \(=\sin 10 x \cdot \frac{\sin ^{10} x}{10}-\frac{10}{10} \int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x=\) \(=\frac{\sin 10 x \cdot \sin ^{10} x}{10}-\int \cos 10 x \sin ^{10} x d x+\int \cos 10 x \cdot \sin ^{10} x d x\) \(I=\frac{\sin 10 x \cdot \sin ^{10} x}{10}+C\)
