Match List I with List II

A-III, B-I, C-II, D-IV

(A) From properties of integration.

If, f(-x) = -f(x) then 

$\int_a^{-a}f(x)dx=0$

f(−x)=sin⁷(−x)

=(−1)⁷sin⁷x

=−sin⁷x ⇒ -f(x) so, \(\int_\frac{-π}{2}^\frac{π}{2}\)sin⁷ xdx = 0

(B) \(\int_{\frac{-π}{2}}^{\frac{π}{2}}\)sin² xdx

\(\int_\frac{-π}{2}^\frac{π}{2}\frac{1-cos2x}{2}dx⇒\begin{bmatrix}\frac{x}{2}-\frac{sin2x}{4}\end{bmatrix}_{\frac{-π}{2}}^{\frac{π}{2}}⇒\frac{π}{2}\)

(C) \(\int_{0}^\frac{π}{2}\frac{1}{1+tan x}dx\)

$=\int_{0}^\frac{π}{2}\frac{dx}{1+tan(\frac{π}{2}-x)}$ (according to property)

$\int_{0}^\frac{π}{2}\frac{dx}{(1-cotx)}=\int_{0}^\frac{π}{2}\frac{tanx}{1+tanx}dx⇒\int_{0}^\frac{π}{2}(1-\frac{1}{tanx})dx=\frac{π}{2}_I$

$I=\frac{π}{4}$

(D) \(\int_{0}^\frac{π}{2}cosxdx\)

$=[sinx]_{0}^\frac{π}{2}⇒[sin\frac{π}{2}-sin0]=1$