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प्रश्न
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उत्तर
\[\int\frac{\tan x}{\sqrt{\cos x}}dx\]
\[ \Rightarrow \int\frac{\sin x}{\cos x \sqrt{\cos x}} dx\]
\[ \Rightarrow \int\frac{\sin x}{\cos {}^\frac{3}{2} x}dx\]
\[Let \cos x = t\]
\[ \Rightarrow - \text{sin x dx }= dt\]
\[ \Rightarrow \sin x = - \frac{dt}{dx}\]
\[Now, \int\frac{\sin x}{\cos {}^\frac{3}{2} x}dx\]
\[ = \int - \frac{1}{t^\frac{3}{2}}dt\]
\[ = - \int t^{- \frac{3}{2}} dt\]
\[ = - \left[ \frac{t^{- \frac{3}{2} + 1}}{\frac{- 3}{2} + 1} \right] + C\]
\[ = \frac{2}{\sqrt{t}} + C\]
\[ = \frac{2}{\sqrt{\cos x}} + C\]
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` = ∫1/{sin^3 x cos^ 2x} dx`
Evaluate the following integrals:
If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]
