Advertisements
Advertisements
प्रश्न
\[\int \sec^4 2x \text{ dx }\]
बेरीज
Advertisements
उत्तर
\[\int \sec^4 2x \text{ dx }\]
∫ sec4 2x dx
= ∫ sec2 2x . sec2 2x dx
= ∫ (1 + tan2 2x) . sec2 2x dx
Let tan 2x = t
⇒ sec2 2x . 2 dx = dt
\[\Rightarrow \sec^2 2x . dx = \frac{dt}{2}\]
\[Now, \int\left( 1 + \tan^2 2x \right) . \sec^2 2x \text{ dx }\]
\[ = \frac{1}{2}\int\left( 1 + t^2 \right) dt\]
\[ = \frac{1}{2}\left[ t + \frac{t^3}{3} \right] + C\]
\[ = \frac{t}{2} + \frac{t^3}{6} + C\]
\[ = \frac{\tan \left( \text{ 2x } \right)}{2} + \frac{\tan^3 \left( \text{ 2x } \right)}{6} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\sin^2 x}{1 + \cos x} \text{dx} \]
\[\int\frac{\tan x}{\sec x + \tan x} dx\]
` ∫ {cosec x} / {"cosec x "- cot x} ` dx
\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]
\[\int\frac{1 + \cos x}{1 - \cos x} dx\]
` ∫ 1/ {1+ cos 3x} ` dx
\[\int\frac{2x + 3}{\left( x - 1 \right)^2} dx\]
\[\int\frac{x + 1}{\sqrt{2x + 3}} dx\]
\[\int\frac{x + 1}{x \left( x + \log x \right)} dx\]
\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]
\[\int\frac{\left( x + 1 \right) e^x}{\sin^2 \left( \text{x e}^x \right)} dx\]
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int\frac{x^2}{\sqrt{3x + 4}} dx\]
` = ∫1/{sin^3 x cos^ 2x} dx`
\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 + 1}} dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{\cos x}{\sqrt{\sin^2 x - 2 \sin x - 3}} dx\]
\[\int\frac{2x}{2 + x - x^2} \text{ dx }\]
\[\int\frac{2x + 1}{\sqrt{x^2 + 2x - 1}}\text{ dx }\]
\[\int\frac{2x + 5}{\sqrt{x^2 + 2x + 5}} dx\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) \text{ dx }\]
\[\int\left( \tan^{- 1} x^2 \right) x\ dx\]
\[\int \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int e^x \cdot \frac{\sqrt{1 - x^2} \sin^{- 1} x + 1}{\sqrt{1 - x^2}} \text{ dx }\]
\[\int\frac{x^2}{\left( x - 1 \right) \left( x - 2 \right) \left( x - 3 \right)} dx\]
\[\int\frac{dx}{\left( x^2 + 1 \right) \left( x^2 + 4 \right)}\]
\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]
\[\int\frac{1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int \sec^6 x\ dx\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int x^3 \left( \log x \right)^2\text{ dx }\]
\[\int\frac{\sin x + \cos x}{\sin^4 x + \cos^4 x} \text{ dx }\]
\[\int e^x \frac{\left( 1 - x \right)^2}{\left( 1 + x^2 \right)^2} \text{ dx }\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
