Question

$\underset{0}{\overset{\frac{\pi }{2}}{\int }}(a{\cos }^{2}x+b{\sin }^{2}x)dx$

Answer 1

Solution :-

$I={\int }_0^{\pi /2}(aco{s}^{2}x+bsi{n}^{2}x)dx$ ___ (1)

$I={\int }_0^{\pi /2}\left(aco{s}^2\right(\frac{\pi }{2}-x)+bsi{n}^2(\frac{\pi }{2}-x\left)\right)dx$

$I={\int }_0^{\pi /2}(asi{n}^{2}x+bco{s}^{2}x)dx$ ___ (2)

Adding (1) and (2)

$2I={\int }_0^{\pi /2}\left(a\right(co{s}^{2}x+si{n}^{2}x)+b(si{n}^{2}x+co{s}^{2}x\left)\right)dx$

$={\int }_0^{\pi /2}(a+b)dx$

$=(a+b)[x{]}_0^{\pi /2}$

$2I=(a+b)\frac{\pi }{2}$

$I=(a+b)\frac{\pi }{4}$