Advertisements
Advertisements
प्रश्न
\[\int \sin^5 x \cos x \text{ dx }\]
बेरीज
Advertisements
उत्तर
∫ sin5 x cos x dx
Let sin x = t
cos x dx = dt
Now, ∫ sin5 x cos x dx
= ∫ t5 . dt
\[= \frac{t^6}{6} + C\]
\[ = \frac{\sin^6 x}{6} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( x + 1 \right)\left( x - 2 \right)}{\sqrt{x}} dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec } {x }- \cot x} dx\]
\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]
\[\int\frac{x^2 + 3x - 1}{\left( x + 1 \right)^2} dx\]
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
\[\int\frac{\tan x}{\sqrt{\cos x}} dx\]
\[\int2x \sec^3 \left( x^2 + 3 \right) \tan \left( x^2 + 3 \right) dx\]
` ∫ e^{m sin ^-1 x}/ \sqrt{1-x^2} ` dx
\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]
\[\int\frac{1}{\sqrt{x} + x} \text{ dx }\]
\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]
\[\int x \cos^3 x^2 \sin x^2 \text{ dx }\]
\[\int\frac{1}{\sin^4 x \cos^2 x} dx\]
\[\int\frac{1}{\sqrt{1 + 4 x^2}} dx\]
\[\int\frac{1}{x^{2/3} \sqrt{x^{2/3} - 4}} dx\]
\[\int\frac{2x - 3}{x^2 + 6x + 13} dx\]
\[\int\frac{x - 1}{\sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{2 \sin x + 3 \cos x}{3 \sin x + 4 \cos x} dx\]
\[\int x \cos x\ dx\]
\[\int x^2 e^{- x} \text{ dx }\]
\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]
\[\int \tan^{- 1} \left( \sqrt{x} \right) \text{dx }\]
\[\int\frac{x \sin^{- 1} x}{\sqrt{1 - x^2}} dx\]
\[\int\left( 4x + 1 \right) \sqrt{x^2 - x - 2} \text{ dx }\]
\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]
Find \[\int\frac{2x}{\left( x^2 + 1 \right) \left( x^2 + 2 \right)^2}dx\]
\[\int\sqrt{\cot \text{θ} d } \text{ θ}\]
\[\int\frac{x^2 - 1}{x^4 + 1} \text{ dx }\]
\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^4 x} dx =\]
\[\int\frac{\cos 2x - 1}{\cos 2x + 1} dx =\]
\[\int\frac{x^9}{\left( 4 x^2 + 1 \right)^6}dx\] is equal to
\[\int\frac{1}{e^x + e^{- x}} dx\]
\[\int \tan^3 x\ dx\]
\[\int \cot^5 x\ dx\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
\[\int\frac{3x + 1}{\sqrt{5 - 2x - x^2}} \text{ dx }\]
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
