Advertisements
Advertisements
प्रश्न
\[\int \sin^3 x \cos^5 x \text{ dx }\]
बेरीज
उत्तर
∫ sin3 x . cos5 x dx
= ∫ sin2 x . cos5 x . sin x dx
= ∫ (1 – cos2 x) . cos5 x sin x dx
Let cos x = t
⇒ – sin x dx = dt
⇒ sin x dx = – dt
Now, ∫ (1 – cos2 x) . cos5 x sin x dx
= –∫ (1 – t2) t5 dt
= –∫ (t5 – t7) dt
= ∫(t7 – t5) dt
\[= \frac{t^8}{8} - \frac{t^6}{6} + C\]
\[ = \frac{\cos^8 x}{8} - \frac{\cos^6 x}{6} + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\left( 2 - 3x \right) \left( 3 + 2x \right) \left( 1 - 2x \right) dx\]
\[\int \tan^{- 1} \left( \frac{\sin 2x}{1 + \cos 2x} \right) dx\]
\[\int\frac{x^3 - 3 x^2 + 5x - 7 + x^2 a^x}{2 x^2} dx\]
\[\int\frac{1}{\sqrt{2x + 3} + \sqrt{2x - 3}} dx\]
\[\int \tan^2 \left( 2x - 3 \right) dx\]
` ∫ sin x \sqrt (1-cos 2x) dx `
` ∫ {sin 2x} /{a cos^2 x + b sin^2 x } ` dx
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
\[\int\left( \frac{x + 1}{x} \right) \left( x + \log x \right)^2 dx\]
\[\int\frac{\cos\sqrt{x}}{\sqrt{x}} dx\]
\[\int \sin^7 x \text{ dx }\]
\[\int\frac{x}{x^4 - x^2 + 1} dx\]
\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]
\[\int\frac{x}{\sqrt{4 - x^4}} dx\]
\[\int\frac{x^2 \left( x^4 + 4 \right)}{x^2 + 4} \text{ dx }\]
\[\int\text{ log }\left( x + 1 \right) \text{ dx }\]
\[\int2 x^3 e^{x^2} dx\]
\[\int x^2 \sin^{- 1} x\ dx\]
\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]
\[\int x^2 \sqrt{a^6 - x^6} \text{ dx}\]
\[\int\left( 2x - 5 \right) \sqrt{2 + 3x - x^2} \text{ dx }\]
\[\int\frac{x^2 + 1}{x\left( x^2 - 1 \right)} dx\]
\[\int\frac{3x + 5}{x^3 - x^2 - x + 1} dx\]
\[\int\frac{3}{\left( 1 - x \right) \left( 1 + x^2 \right)} dx\]
Evaluate the following integral:
\[\int\frac{x^2}{1 - x^4}dx\]
\[\int\frac{x + 1}{\left( x - 1 \right) \sqrt{x + 2}} \text{ dx }\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]
If \[\int\frac{2^{1/x}}{x^2} dx = k 2^{1/x} + C,\] then k is equal to
\[\int\sqrt{\frac{x}{1 - x}} dx\] is equal to
\[\int \sec^2 x \cos^2 2x \text{ dx }\]
\[\int \tan^5 x\ dx\]
\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]
\[\int \sin^5 x\ dx\]
\[\int\sqrt{\text{ cosec x} - 1} \text{ dx }\]
\[\int\frac{\sin^2 x}{\cos^6 x} \text{ dx }\]
\[\int \left( \sin^{- 1} x \right)^3 dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} \text{ dx}\]
\[\int\frac{5 x^4 + 12 x^3 + 7 x^2}{x^2 + x} dx\]
Find: `int (3x +5)/(x^2+3x-18)dx.`