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Evaluate the Following Integral: ∫ π 2 0 a Sin X + B Sin X Sin X + Cos X D X - Mathematics

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प्रश्न

Evaluate the following integral:

\[\int_0^\frac{\pi}{2} \frac{a\sin x + b\sin x}{\sin x + \cos x}dx\]

 

योग

उत्तर

\[\text{Let I }=\int_0^\frac{\pi}{2} \frac{a\sin x + b\sin x}{\sin x + \cos x}dx...............(1)\]

Then,

\[I = \int_0^\frac{\pi}{2} \frac{a\sin\left( \frac{\pi}{2} - x \right) + b\cos\left( \frac{\pi}{2} - x \right)}{\sin\left( \frac{\pi}{2} - x \right) + \cos\left( \frac{\pi}{2} - x \right)}dx ...................\left[ \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]

\[= \int_0^\frac{\pi}{2} \frac{a\cos x + b\sin x}{\cos x + \sin x}dx................(2)\]

Adding (1) and (2), we get

\[2I = \int_0^\frac{\pi}{2} \left( \frac{a\sin x + b\cos x}{\cos x + \sin x} + \frac{a\cos x + b\sin x}{\sin x + \cos x} \right)dx\]
\[ \Rightarrow 2I = \int_0^\frac{\pi}{2} \left( \frac{a\sin x + b\cos x + a\cos x + b\sin x}{\sin x + \cos x} \right)dx\]
\[ \Rightarrow 2I = \int_0^\frac{\pi}{2} \frac{\left( a + b \right)\sin x + \left( a + b \right)\cos x}{\sin x + \cos x}dx\]
\[ \Rightarrow 2I = \int_0^\frac{\pi}{2} \frac{\left( a + b \right)\left( \sin x + \cos x \right)}{\sin x + \cos x}dx\]

\[\Rightarrow 2I = \int_0^\frac{\pi}{2} \left( a + b \right)dx\]
\[ \Rightarrow 2I = \left( a + b \right) \times \left.x\right|_0^\frac{\pi}{2} \]
\[ \Rightarrow 2I = \left( a + b \right) \times \left( \frac{\pi}{2} - 0 \right)\]
\[ \Rightarrow 2I = \frac{\pi}{2}\left( a + b \right)\]
\[ \Rightarrow I = \frac{\pi}{4}\left( a + b \right)\]

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अध्याय 20: Definite Integrals - Exercise 20.5 [पृष्ठ ९५]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.5 | Q 39 | पृष्ठ ९५

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