(B) : Given, \(I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x\) \(I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x}\left[\frac{\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x\) \(I=\int e^{x}\left(\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right) d x\) Let, \(\quad f(x)=\frac{1}{1+x^{2}}\) \(f^{\prime}(x)=\frac{-2 x}{(1+x)^{2}}\) The given function is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C\) So, \(\quad \mathrm{I}=\mathrm{e}^{\mathrm{x}}\left(\frac{1}{1+\mathrm{x}^{2}}\right)+\mathrm{C}\)
Kerala CEE-2009
Integral Calculus
86279
If \(I(x)=I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) and \(I(0)=1\), then \(I\left(\frac{\pi}{3}\right)\) is equal to
1 \(-\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
2 \(\mathrm{e}^{\frac{3}{4}}\)
3 \(\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
4 \(-\mathrm{e}^{\frac{3}{4}}\)
Explanation:
(C) : Given, \(I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) \(=\int \mathrm{e}^{\sin ^{2} x} \cdot \cos \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx}-\int \mathrm{e}^{\sin ^{2} x} \cdot \sin \mathrm{x} \mathrm{dx}\) \(=\int(\cos x) \cdot\left(e^{\sin ^{2} x} \cdot \sin 2 x\right) d x-\int e^{\sin ^{2} x} \cdot \sin x d x\) \(=\cos x \cdot e^{\sin ^{2} x}-\int(-\sin x) e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(=\cos x \cdot e^{\sin ^{2} x}+\int \sin x \cdot e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\because\) Given, \(\mathrm{I}(0)=1\) \(\mathrm{c}=0\) So, \(\quad I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\mathrm{I}\left(\frac{\pi}{3}\right)=\mathrm{e}^{\sin ^{2}\left(\frac{\pi}{3}\right)} \cdot \cos \left(\frac{\pi}{3}\right)\) \(=\mathrm{e}^{\frac{3}{4}} \cdot \frac{1}{2}=\frac{\mathrm{e}^{\frac{3}{4}}}{2}\)
(A) : Let \(\mathrm{x}=\mathrm{t}^{2}, \mathrm{dx}=2 \mathrm{t} \mathrm{dt}\) \(I=2 \int(\cos t) \cdot t d t\) \(=2 t \cdot \int \cos t d t-\int\left[\frac{d}{d t} t \int \cos t d t\right] d t\) \(I=2\left[t \cdot \int \cos t d t-\int(\sin t) d t\right]\) \(I=2[t \sin t+\cos t]+c\) \(I=2 \sqrt{x} \sin \sqrt{x}+2 \cos \sqrt{x}+c\)
APEAPCET-2021-20.08.2021
Integral Calculus
86281
If \(\int \cdot\left[\cos (x) \cdot \frac{d}{d x}(\operatorname{cosec}(x)] d x=f(x)+g(x)+\mathbf{c}\right.\), then \(\mathbf{f}(\mathbf{x}) . \mathbf{g}(\mathbf{x})=\)
(B) : Given, \(I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x\) \(I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x}\left[\frac{\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x\) \(I=\int e^{x}\left(\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right) d x\) Let, \(\quad f(x)=\frac{1}{1+x^{2}}\) \(f^{\prime}(x)=\frac{-2 x}{(1+x)^{2}}\) The given function is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C\) So, \(\quad \mathrm{I}=\mathrm{e}^{\mathrm{x}}\left(\frac{1}{1+\mathrm{x}^{2}}\right)+\mathrm{C}\)
Kerala CEE-2009
Integral Calculus
86279
If \(I(x)=I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) and \(I(0)=1\), then \(I\left(\frac{\pi}{3}\right)\) is equal to
1 \(-\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
2 \(\mathrm{e}^{\frac{3}{4}}\)
3 \(\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
4 \(-\mathrm{e}^{\frac{3}{4}}\)
Explanation:
(C) : Given, \(I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) \(=\int \mathrm{e}^{\sin ^{2} x} \cdot \cos \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx}-\int \mathrm{e}^{\sin ^{2} x} \cdot \sin \mathrm{x} \mathrm{dx}\) \(=\int(\cos x) \cdot\left(e^{\sin ^{2} x} \cdot \sin 2 x\right) d x-\int e^{\sin ^{2} x} \cdot \sin x d x\) \(=\cos x \cdot e^{\sin ^{2} x}-\int(-\sin x) e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(=\cos x \cdot e^{\sin ^{2} x}+\int \sin x \cdot e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\because\) Given, \(\mathrm{I}(0)=1\) \(\mathrm{c}=0\) So, \(\quad I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\mathrm{I}\left(\frac{\pi}{3}\right)=\mathrm{e}^{\sin ^{2}\left(\frac{\pi}{3}\right)} \cdot \cos \left(\frac{\pi}{3}\right)\) \(=\mathrm{e}^{\frac{3}{4}} \cdot \frac{1}{2}=\frac{\mathrm{e}^{\frac{3}{4}}}{2}\)
(A) : Let \(\mathrm{x}=\mathrm{t}^{2}, \mathrm{dx}=2 \mathrm{t} \mathrm{dt}\) \(I=2 \int(\cos t) \cdot t d t\) \(=2 t \cdot \int \cos t d t-\int\left[\frac{d}{d t} t \int \cos t d t\right] d t\) \(I=2\left[t \cdot \int \cos t d t-\int(\sin t) d t\right]\) \(I=2[t \sin t+\cos t]+c\) \(I=2 \sqrt{x} \sin \sqrt{x}+2 \cos \sqrt{x}+c\)
APEAPCET-2021-20.08.2021
Integral Calculus
86281
If \(\int \cdot\left[\cos (x) \cdot \frac{d}{d x}(\operatorname{cosec}(x)] d x=f(x)+g(x)+\mathbf{c}\right.\), then \(\mathbf{f}(\mathbf{x}) . \mathbf{g}(\mathbf{x})=\)
(B) : Given, \(I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x\) \(I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x}\left[\frac{\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x\) \(I=\int e^{x}\left(\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right) d x\) Let, \(\quad f(x)=\frac{1}{1+x^{2}}\) \(f^{\prime}(x)=\frac{-2 x}{(1+x)^{2}}\) The given function is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C\) So, \(\quad \mathrm{I}=\mathrm{e}^{\mathrm{x}}\left(\frac{1}{1+\mathrm{x}^{2}}\right)+\mathrm{C}\)
Kerala CEE-2009
Integral Calculus
86279
If \(I(x)=I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) and \(I(0)=1\), then \(I\left(\frac{\pi}{3}\right)\) is equal to
1 \(-\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
2 \(\mathrm{e}^{\frac{3}{4}}\)
3 \(\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
4 \(-\mathrm{e}^{\frac{3}{4}}\)
Explanation:
(C) : Given, \(I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) \(=\int \mathrm{e}^{\sin ^{2} x} \cdot \cos \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx}-\int \mathrm{e}^{\sin ^{2} x} \cdot \sin \mathrm{x} \mathrm{dx}\) \(=\int(\cos x) \cdot\left(e^{\sin ^{2} x} \cdot \sin 2 x\right) d x-\int e^{\sin ^{2} x} \cdot \sin x d x\) \(=\cos x \cdot e^{\sin ^{2} x}-\int(-\sin x) e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(=\cos x \cdot e^{\sin ^{2} x}+\int \sin x \cdot e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\because\) Given, \(\mathrm{I}(0)=1\) \(\mathrm{c}=0\) So, \(\quad I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\mathrm{I}\left(\frac{\pi}{3}\right)=\mathrm{e}^{\sin ^{2}\left(\frac{\pi}{3}\right)} \cdot \cos \left(\frac{\pi}{3}\right)\) \(=\mathrm{e}^{\frac{3}{4}} \cdot \frac{1}{2}=\frac{\mathrm{e}^{\frac{3}{4}}}{2}\)
(A) : Let \(\mathrm{x}=\mathrm{t}^{2}, \mathrm{dx}=2 \mathrm{t} \mathrm{dt}\) \(I=2 \int(\cos t) \cdot t d t\) \(=2 t \cdot \int \cos t d t-\int\left[\frac{d}{d t} t \int \cos t d t\right] d t\) \(I=2\left[t \cdot \int \cos t d t-\int(\sin t) d t\right]\) \(I=2[t \sin t+\cos t]+c\) \(I=2 \sqrt{x} \sin \sqrt{x}+2 \cos \sqrt{x}+c\)
APEAPCET-2021-20.08.2021
Integral Calculus
86281
If \(\int \cdot\left[\cos (x) \cdot \frac{d}{d x}(\operatorname{cosec}(x)] d x=f(x)+g(x)+\mathbf{c}\right.\), then \(\mathbf{f}(\mathbf{x}) . \mathbf{g}(\mathbf{x})=\)
(B) : Given, \(I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x\) \(I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x}\left[\frac{\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x\) \(I=\int e^{x}\left(\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right) d x\) Let, \(\quad f(x)=\frac{1}{1+x^{2}}\) \(f^{\prime}(x)=\frac{-2 x}{(1+x)^{2}}\) The given function is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C\) So, \(\quad \mathrm{I}=\mathrm{e}^{\mathrm{x}}\left(\frac{1}{1+\mathrm{x}^{2}}\right)+\mathrm{C}\)
Kerala CEE-2009
Integral Calculus
86279
If \(I(x)=I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) and \(I(0)=1\), then \(I\left(\frac{\pi}{3}\right)\) is equal to
1 \(-\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
2 \(\mathrm{e}^{\frac{3}{4}}\)
3 \(\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
4 \(-\mathrm{e}^{\frac{3}{4}}\)
Explanation:
(C) : Given, \(I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) \(=\int \mathrm{e}^{\sin ^{2} x} \cdot \cos \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx}-\int \mathrm{e}^{\sin ^{2} x} \cdot \sin \mathrm{x} \mathrm{dx}\) \(=\int(\cos x) \cdot\left(e^{\sin ^{2} x} \cdot \sin 2 x\right) d x-\int e^{\sin ^{2} x} \cdot \sin x d x\) \(=\cos x \cdot e^{\sin ^{2} x}-\int(-\sin x) e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(=\cos x \cdot e^{\sin ^{2} x}+\int \sin x \cdot e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\because\) Given, \(\mathrm{I}(0)=1\) \(\mathrm{c}=0\) So, \(\quad I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\mathrm{I}\left(\frac{\pi}{3}\right)=\mathrm{e}^{\sin ^{2}\left(\frac{\pi}{3}\right)} \cdot \cos \left(\frac{\pi}{3}\right)\) \(=\mathrm{e}^{\frac{3}{4}} \cdot \frac{1}{2}=\frac{\mathrm{e}^{\frac{3}{4}}}{2}\)
(A) : Let \(\mathrm{x}=\mathrm{t}^{2}, \mathrm{dx}=2 \mathrm{t} \mathrm{dt}\) \(I=2 \int(\cos t) \cdot t d t\) \(=2 t \cdot \int \cos t d t-\int\left[\frac{d}{d t} t \int \cos t d t\right] d t\) \(I=2\left[t \cdot \int \cos t d t-\int(\sin t) d t\right]\) \(I=2[t \sin t+\cos t]+c\) \(I=2 \sqrt{x} \sin \sqrt{x}+2 \cos \sqrt{x}+c\)
APEAPCET-2021-20.08.2021
Integral Calculus
86281
If \(\int \cdot\left[\cos (x) \cdot \frac{d}{d x}(\operatorname{cosec}(x)] d x=f(x)+g(x)+\mathbf{c}\right.\), then \(\mathbf{f}(\mathbf{x}) . \mathbf{g}(\mathbf{x})=\)
(B) : Given, \(I=\int e^{x}\left(\frac{1-x}{1+x^{2}}\right)^{2} d x\) \(I=\int e^{x} \frac{(1-x)^{2}}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x} \frac{\left(1+x^{2}-2 x\right)}{\left(1+x^{2}\right)^{2}} d x\) \(I=\int e^{x}\left[\frac{\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right] d x\) \(I=\int e^{x}\left(\frac{1}{1+x^{2}}-\frac{2 x}{\left(1+x^{2}\right)^{2}}\right) d x\) Let, \(\quad f(x)=\frac{1}{1+x^{2}}\) \(f^{\prime}(x)=\frac{-2 x}{(1+x)^{2}}\) The given function is of the form of \(\int e^{x}\left[f(x)+f^{\prime}(x)\right] d x=e^{x} f(x)+C\) So, \(\quad \mathrm{I}=\mathrm{e}^{\mathrm{x}}\left(\frac{1}{1+\mathrm{x}^{2}}\right)+\mathrm{C}\)
Kerala CEE-2009
Integral Calculus
86279
If \(I(x)=I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) and \(I(0)=1\), then \(I\left(\frac{\pi}{3}\right)\) is equal to
1 \(-\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
2 \(\mathrm{e}^{\frac{3}{4}}\)
3 \(\frac{1}{2} \mathrm{e}^{\frac{3}{4}}\)
4 \(-\mathrm{e}^{\frac{3}{4}}\)
Explanation:
(C) : Given, \(I(x)=\int e^{\sin ^{2} x} \cdot(\cos x \cdot \sin 2 x-\sin x) d x\) \(=\int \mathrm{e}^{\sin ^{2} x} \cdot \cos \mathrm{x} \cdot \sin 2 \mathrm{x} \mathrm{dx}-\int \mathrm{e}^{\sin ^{2} x} \cdot \sin \mathrm{x} \mathrm{dx}\) \(=\int(\cos x) \cdot\left(e^{\sin ^{2} x} \cdot \sin 2 x\right) d x-\int e^{\sin ^{2} x} \cdot \sin x d x\) \(=\cos x \cdot e^{\sin ^{2} x}-\int(-\sin x) e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(=\cos x \cdot e^{\sin ^{2} x}+\int \sin x \cdot e^{\sin ^{2} x} d x-\int \sin x \cdot e^{\sin ^{2} x} d x\) \(I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\because\) Given, \(\mathrm{I}(0)=1\) \(\mathrm{c}=0\) So, \(\quad I(x)=e^{\sin ^{2} x} \cdot \cos x\) \(\mathrm{I}\left(\frac{\pi}{3}\right)=\mathrm{e}^{\sin ^{2}\left(\frac{\pi}{3}\right)} \cdot \cos \left(\frac{\pi}{3}\right)\) \(=\mathrm{e}^{\frac{3}{4}} \cdot \frac{1}{2}=\frac{\mathrm{e}^{\frac{3}{4}}}{2}\)
(A) : Let \(\mathrm{x}=\mathrm{t}^{2}, \mathrm{dx}=2 \mathrm{t} \mathrm{dt}\) \(I=2 \int(\cos t) \cdot t d t\) \(=2 t \cdot \int \cos t d t-\int\left[\frac{d}{d t} t \int \cos t d t\right] d t\) \(I=2\left[t \cdot \int \cos t d t-\int(\sin t) d t\right]\) \(I=2[t \sin t+\cos t]+c\) \(I=2 \sqrt{x} \sin \sqrt{x}+2 \cos \sqrt{x}+c\)
APEAPCET-2021-20.08.2021
Integral Calculus
86281
If \(\int \cdot\left[\cos (x) \cdot \frac{d}{d x}(\operatorname{cosec}(x)] d x=f(x)+g(x)+\mathbf{c}\right.\), then \(\mathbf{f}(\mathbf{x}) . \mathbf{g}(\mathbf{x})=\)