Match List I with List II
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A.\(\int_\frac{-π}{2}^\frac{π}{2}\)sin7 xdx |
I.\(\frac{π}{2}\) |
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B.\(\int_\frac{-π}{2}^\frac{π}{2}\)sin2 xdx |
II.\(\frac{π}{4}\) |
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C.\(\int_{0}^\frac{π}{2}d\)\(\frac{1}{1+tan x}\)dx |
III. 0 |
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D.\(\int_{0}^\frac{π}{2}\)\(\left|cosx\right|\) |
IV. 2
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Answer & explanation
Correct answer: option 2
(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) \(\int_{0}^\frac{π}{2}cosxdx\)
$=[sinx]_{0}^\frac{π}{2}⇒[sin\frac{π}{2}-sin0]=1$