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Integrals

Multiple Choice Questions

71.

The value of $\int_{- 2}^{2} (xcos (x) + \sin (x) + 1) d x$

72.

$\int_{π}^{16 π} |\sin (x)| d x$ is equal to

73.

$\int \frac{\cos (2 x)}{\cos (x)} dx$ is equal to

$2 \sin (x) + \log |\sec (x) + \tan (x)| + C$
$2 \sin (x) - \log |\sec (x) - \tan (x)| + C$
$2 \sin (x) - \log |\sec (x) + \tan (x)| + C$
$2 \sin (x) + \log |\sec (x) - \tan (x)| + C$

74.

$\int \frac{\sin^{8} (x) - \cos^{8} (x)}{1 - 2 \sin^{2} (x) \cos^{2} (x)} dx$

$- \frac{1}{2} \sin (2 x) + C$
$\frac{1}{2} \sin (2 x) + C$
$\frac{1}{2} \sin (x) + C$
$- \frac{1}{2} \sin (x) + C$

75.

The value of $\int_{0}^{π} \sin^{50} (x) \cos^{49} (x) d x$ is

0
$\frac{π}{4}$
$\frac{π}{2}$
1

76.

$\int 2^{x} (f' (x) + f (x) \log (2)) dx$ is

2^xf'(x) + C
2^xf(x) + C
2^x(log(2))f(x) + C
log(2)f(x) + C

Short Answer Type

77.
Evaluate the following integral
$\int_{- 1}^{2} |xsin (πx)| d x$

$I = \int_{- 1}^{2} |xsin (πx)| d x = \int_{- 1}^{1} |xsin (πx)| d x = \int_{1}^{2} |xsin (πx)| d x = 2 \int_{0}^{1} xsin (πx) d x + \int_{1}^{2} xsin (πx) d x = 2 \int_{0}^{1} xsin (πx) d x - \int_{1}^{2} xsin (πx) d x = 2 I_{1} - I_{2} I_{1} = \int_{0}^{1} xsin (πx) d x = {[- x \frac{\cos (πx)}{π}]}_{0}^{1} + \int_{0}^{1} \frac{\cos (πx)}{π} d x = {[- x \frac{\cos (πx)}{π} + \frac{\sin (πx)}{π^{2}}]}_{0}^{1} = \frac{1}{π} and I_{2} = \int_{1}^{2} xsin (πx) d x = {[- x \frac{\cos (πx)}{π} + \frac{\sin (πx)}{π^{2}}]}_{1}^{2} = \frac{- 2}{π} + 0 + (- \frac{1}{π}) = - \frac{3}{π} So, 2 I_{1} - I_{2} = \frac{2}{π} + \frac{3}{π} = \frac{5}{π} \Rightarrow \int_{- 1}^{2} |xsin (πx)| d x = \frac{5}{π}$