MCQ
The value of $\int_a^{a + (\pi /2)} {({{\sin }^4}x + {{\cos }^4}x)\,dx} $ is
- AIndependent of $a$
- B$a\,{\left( {\frac{\pi }{2}} \right)^2}$
- ✓$\frac{{3\pi }}{8}$
- D$\frac{{3\pi {a^2}}}{8}$
therefore $\int_a^{a + (\pi /2)} {({{\sin }^4}x + {{\cos }^4}x){\rm{ }}dx} $
$ = \int_0^{\pi /2} {({{\sin }^4}x + {{\cos }^4}x)dx} $
$ = 2\int_0^{\pi /2} {{{\sin }^4}x\,dx = \frac{{3\Gamma (5/2)\Gamma (1/2)}}{{2\Gamma \left( {\frac{{4 + 0 + 2}}{2}} \right)}} = \frac{{3\pi }}{8}} $.
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