मराठी

Π / 2 ∫ 0 1 1 + Tan 3 X D X

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प्रश्न

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]

बेरीज
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उत्तर

\[Let, I = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 x} d x ..............(1)\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 \left( \frac{\pi}{2} - x \right)} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^3 x} d x ................(2)\]

Adding (1) and (2)

\[2I = \int_0^\frac{\pi}{2} \left[ \frac{1}{1 + \tan^3 x} + \frac{1}{1 + co t^3 x} \right] d x\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan^3 x + co t^3 x}{\left( 1 + \tan^3 x \right)\left( 1 + co t^3 x \right)}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan^3 x + co t^3 x}{2 + \tan^3 x + co t^3 x}dx\]

\[ = \int_0^\frac{\pi}{2} dx \]

\[ = \left( x \right)_0^\frac{\pi}{2} \]

\[ = \frac{\pi}{2}\]

\[Hence, I = \frac{\pi}{4}\]

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पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 40 | पृष्ठ १२२

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