86710
\(\int_{0}^{2 \pi} \sin ^{9} x d x\) is equal to
1 0
2 \(18 \pi\)
3 \(9 \pi\)
4 18
Explanation:
(A) : Let, \(\mathrm{I}=\int_{0}^{2 \pi} \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int_{0}^{2 \pi} \sin ^{9}(2 \pi-x) d x\) \(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(I=\int_{0}^{2 \pi}-\sin ^{9} x d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I} =0\) \(\mathrm{I} =0\)
[JCECE-2018]
Integral Calculus
86711
The value of the integral \(\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x, a>0\) is
1 1
2 0
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(C) : Let, \(\mathrm{I}=\int_{-\pi}^{\pi} \frac{\cos ^{2} \mathrm{x}}{1+\mathrm{a}^{\mathrm{x}}} \mathrm{dx}\) \(=\int_{-\pi}^{\pi} \frac{\cos ^{2}(-x)}{1+a^{-x}} d x=\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{-x}} d x\) \(I=\int_{-\pi}^{\pi} \frac{a^{x} \cos ^{2} x}{a^{x}+1} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{-\pi}^{\pi} \cos ^{2} x d x=2 \int_{0}^{\pi} \cos ^{2} x d x\) \(=2 \times 2 \int_{0}^{\pi / 2} \cos ^{2} \mathrm{xdx} \quad[\because \mathrm{f}(\pi-\mathrm{x})=\mathrm{f}(\mathrm{x})]\) \(=4 \int_{0}^{\pi / 2} \frac{1+\cos 2 x}{2} d x=2\left[x+\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=2\left[\frac{\pi}{2}\right]=\pi\) \(\mathrm{I}=\frac{\pi}{2}\)
86710
\(\int_{0}^{2 \pi} \sin ^{9} x d x\) is equal to
1 0
2 \(18 \pi\)
3 \(9 \pi\)
4 18
Explanation:
(A) : Let, \(\mathrm{I}=\int_{0}^{2 \pi} \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int_{0}^{2 \pi} \sin ^{9}(2 \pi-x) d x\) \(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(I=\int_{0}^{2 \pi}-\sin ^{9} x d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I} =0\) \(\mathrm{I} =0\)
[JCECE-2018]
Integral Calculus
86711
The value of the integral \(\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x, a>0\) is
1 1
2 0
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(C) : Let, \(\mathrm{I}=\int_{-\pi}^{\pi} \frac{\cos ^{2} \mathrm{x}}{1+\mathrm{a}^{\mathrm{x}}} \mathrm{dx}\) \(=\int_{-\pi}^{\pi} \frac{\cos ^{2}(-x)}{1+a^{-x}} d x=\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{-x}} d x\) \(I=\int_{-\pi}^{\pi} \frac{a^{x} \cos ^{2} x}{a^{x}+1} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{-\pi}^{\pi} \cos ^{2} x d x=2 \int_{0}^{\pi} \cos ^{2} x d x\) \(=2 \times 2 \int_{0}^{\pi / 2} \cos ^{2} \mathrm{xdx} \quad[\because \mathrm{f}(\pi-\mathrm{x})=\mathrm{f}(\mathrm{x})]\) \(=4 \int_{0}^{\pi / 2} \frac{1+\cos 2 x}{2} d x=2\left[x+\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=2\left[\frac{\pi}{2}\right]=\pi\) \(\mathrm{I}=\frac{\pi}{2}\)
86710
\(\int_{0}^{2 \pi} \sin ^{9} x d x\) is equal to
1 0
2 \(18 \pi\)
3 \(9 \pi\)
4 18
Explanation:
(A) : Let, \(\mathrm{I}=\int_{0}^{2 \pi} \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int_{0}^{2 \pi} \sin ^{9}(2 \pi-x) d x\) \(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(I=\int_{0}^{2 \pi}-\sin ^{9} x d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I} =0\) \(\mathrm{I} =0\)
[JCECE-2018]
Integral Calculus
86711
The value of the integral \(\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x, a>0\) is
1 1
2 0
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(C) : Let, \(\mathrm{I}=\int_{-\pi}^{\pi} \frac{\cos ^{2} \mathrm{x}}{1+\mathrm{a}^{\mathrm{x}}} \mathrm{dx}\) \(=\int_{-\pi}^{\pi} \frac{\cos ^{2}(-x)}{1+a^{-x}} d x=\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{-x}} d x\) \(I=\int_{-\pi}^{\pi} \frac{a^{x} \cos ^{2} x}{a^{x}+1} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{-\pi}^{\pi} \cos ^{2} x d x=2 \int_{0}^{\pi} \cos ^{2} x d x\) \(=2 \times 2 \int_{0}^{\pi / 2} \cos ^{2} \mathrm{xdx} \quad[\because \mathrm{f}(\pi-\mathrm{x})=\mathrm{f}(\mathrm{x})]\) \(=4 \int_{0}^{\pi / 2} \frac{1+\cos 2 x}{2} d x=2\left[x+\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=2\left[\frac{\pi}{2}\right]=\pi\) \(\mathrm{I}=\frac{\pi}{2}\)
86710
\(\int_{0}^{2 \pi} \sin ^{9} x d x\) is equal to
1 0
2 \(18 \pi\)
3 \(9 \pi\)
4 18
Explanation:
(A) : Let, \(\mathrm{I}=\int_{0}^{2 \pi} \sin ^{9} \mathrm{x} d \mathrm{x}\) \(I=\int_{0}^{2 \pi} \sin ^{9}(2 \pi-x) d x\) \(\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]\) \(I=\int_{0}^{2 \pi}-\sin ^{9} x d x\) Adding equation (i) and (ii), we get - \(2 \mathrm{I} =0\) \(\mathrm{I} =0\)
[JCECE-2018]
Integral Calculus
86711
The value of the integral \(\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{x}} d x, a>0\) is
1 1
2 0
3 \(\frac{\pi}{2}\)
4 \(\pi\)
Explanation:
(C) : Let, \(\mathrm{I}=\int_{-\pi}^{\pi} \frac{\cos ^{2} \mathrm{x}}{1+\mathrm{a}^{\mathrm{x}}} \mathrm{dx}\) \(=\int_{-\pi}^{\pi} \frac{\cos ^{2}(-x)}{1+a^{-x}} d x=\int_{-\pi}^{\pi} \frac{\cos ^{2} x}{1+a^{-x}} d x\) \(I=\int_{-\pi}^{\pi} \frac{a^{x} \cos ^{2} x}{a^{x}+1} d x\) Adding equation (i) and (ii), we get- \(2 I=\int_{-\pi}^{\pi} \cos ^{2} x d x=2 \int_{0}^{\pi} \cos ^{2} x d x\) \(=2 \times 2 \int_{0}^{\pi / 2} \cos ^{2} \mathrm{xdx} \quad[\because \mathrm{f}(\pi-\mathrm{x})=\mathrm{f}(\mathrm{x})]\) \(=4 \int_{0}^{\pi / 2} \frac{1+\cos 2 x}{2} d x=2\left[x+\frac{\sin 2 x}{2}\right]_{0}^{\pi / 2}=2\left[\frac{\pi}{2}\right]=\pi\) \(\mathrm{I}=\frac{\pi}{2}\)
