Advertisements
Advertisements
प्रश्न
` ∫ sec^6 x tan x dx `
योग
Advertisements
उत्तर
` ∫ sec^6 x tan x dx `
=∫ sec6 x.sec x tan x dx
Let sec x = t
Let sec x = t
⇒ sec x tan x dx = dt
Now, ∫ sec6 x.sec x tan x dx
= ∫ t6. dt
Now, ∫ sec6 x.sec x tan x dx
= ∫ t6. dt
\[= \frac{t^6}{6} + C\]
\[ = \frac{\sec^6 x}{6} + C\]
\[ = \frac{\sec^6 x}{6} + C\]
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
APPEARS IN
संबंधित प्रश्न
\[\int\left\{ x^2 + e^{\log x}+ \left( \frac{e}{2} \right)^x \right\} dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
` ∫ {"cosec" x }/ { log tan x/2 ` dx
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\left( 4x + 2 \right)\sqrt{x^2 + x + 1} \text{dx}\]
\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
` ∫ x {tan^{- 1} x^2}/{1 + x^4} dx`
\[\int\frac{e^{2x}}{1 + e^x} dx\]
\[\int 5^{5^{5^x}} 5^{5^x} 5^x dx\]
\[\int {cosec}^4 \text{ 3x } \text{ dx } \]
\[\int \sin^5 x \cos x \text{ dx }\]
\[\int \cos^7 x \text{ dx } \]
Evaluate the following integrals:
\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]
` ∫ \sqrt{"cosec x"- 1} dx `
\[\int\frac{x + 2}{\sqrt{x^2 - 1}} \text{ dx }\]
\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]
\[\int\frac{1}{1 - \tan x} \text{ dx }\]
\[\int x e^{2x} \text{ dx }\]
\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2} \text{ dx }\]
` ∫ x tan ^2 x dx
\[\int x^2 \tan^{- 1} x\text{ dx }\]
\[\int \cos^3 \sqrt{x}\ dx\]
\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx }\]
\[\int e^x \left( \log x + \frac{1}{x} \right) dx\]
∴\[\int e^{2x} \left( - \sin x + 2 \cos x \right) dx\]
\[\int\frac{x^2 + 1}{x^2 - 1} dx\]
\[\int\frac{2 x^2 + 7x - 3}{x^2 \left( 2x + 1 \right)} dx\]
\[\int\frac{1}{x \left( x^4 + 1 \right)} dx\]
\[\int\frac{1}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)} dx\]
\[\int\frac{\left( x^2 + 1 \right) \left( x^2 + 2 \right)}{\left( x^2 + 3 \right) \left( x^2 + 4 \right)} dx\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int \cos^3 (3x)\ dx\]
\[\int\sin x \sin 2x \text{ sin 3x dx }\]
\[\int\frac{1}{5 - 4 \sin x} \text{ dx }\]
\[\int \tan^5 x\ \sec^3 x\ dx\]
\[\int \left( x + 1 \right)^2 e^x \text{ dx }\]
\[\int\frac{1}{1 + x + x^2 + x^3} \text{ dx }\]
