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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\sin\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\ ...(\text{i})$
Then,
$\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sin\big(\frac{\pi}{2}-\text{x}\big)+\text{b}\cos\big(\frac{\pi}{2}-\text{x}\big)}{\sin\big(\frac{\pi}{2}-\text{x}\big)+\cos\big(\frac{\pi}{2}-\text{x}\big)}\text{ dx}$ $\Bigg[\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Bigg]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\cos\text{x}+\sin\text{x}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii) we get
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}}{\cos\text{x}+\sin\text{x}}+\frac{\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\Big(\frac{\text{a}\sin\text{x}+\text{b}\cos\text{x}+\text{a}\cos\text{x}+\text{b}\sin\text{x}}{\sin\text{x}+\cos\text{x}}\Big)\text{dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})\sin\text{x}+(\text{a}+\text{b})\cos\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{(\text{a}+\text{b})(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx}$
$\Rightarrow2\text{I}=\int\limits^{\frac{\pi}{2}}_0(\text{a}+\text{b})\text{dx}$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\big[\text{x}\big]^{\frac{\pi}{2}}_0$
$\Rightarrow2\text{I}=(\text{a}+\text{b})\times\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow2\text{I}=\frac{\pi}{2}(\text{a}+\text{b})$
$\Rightarrow\text{I}=\frac{\pi}{4}(\text{a}+\text{b})$

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