A) \[-\frac{1}{2}{{\sin }^{2}}x+c\]
B) \[\frac{1}{2}{{\sin }^{2}}x+c\]
C) \[-\frac{1}{2}\sin {{x}^{2}}+c\]
D) \[\frac{1}{2}\sin {{x}^{2}}+c\]
Correct Answer: D
Solution :
Put \[{{x}^{2}}=t\Rightarrow dt=2x\,dx\] \ Given integral\[=\frac{1}{2}\int_{{}}^{{}}{\cos t\,dt}=\frac{1}{2}\sin t=\frac{1}{2}\sin {{x}^{2}}+c\].
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