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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Prove that $\int\limits_{0}^{\text{a}}\text{f(x)dx}=\int\limits_{0}^{\text{a}}\text{f (a - x) dx.}$
Hence, evaluate $\int\limits_{0}^{\pi/2}\frac{\text{dx}}{\text{1 + tan x}}$.
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Answer

$\text{RHS}=\int\limits^{a}_{0}\text{f (a - x)dx}=-\int\limits^{a}_{0}\text{f (y)dy }\text{ }\text{ where (a - x)}=\text{y} $
$=\int\limits_{0}^{a}\text{f (y) dy}=\int\limits_{0}^{a}\text{f (x) dx}=\text{LHS}$
$\text{I}=\int\limits_{0}^{\pi/2}\frac{\text{dx}}{\text{1 + tan x}}=\int\limits_{0}^{\pi/2}\frac{\cos\text{x}}{\cos\text{x + sin x}}\text{dx}...........\text{(i)}$
$=\int\limits_{0}^{\pi/2}\frac{\cos\text{ }(\pi/2-\text{x})}{\cos(\pi/2-\text{x})+\sin(\pi/2-\text{x})}\text{dx}=\int\limits_{0}^{\pi/2}\frac{\sin\text{ x}}{\text{sin x + cos x}}\text{dx}...........\text{(ii)}$
$\Rightarrow\text{2I}=\int_{0}^{\pi/2}\text{1 dx}=[\text{x}]_0^{\pi/2}=\pi/2$
$\Rightarrow\text{I}=\pi/4.$

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