मराठी

∫ Sin 5 X Cos X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int \sin^5 x \cos x \text{ dx }\]
बेरीज

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?
पाठ 19: Indefinite Integrals - Exercise 19.12 [पृष्ठ ७३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.12 | Q 4 | पृष्ठ ७३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\frac{\left( 1 + \sqrt{x} \right)^2}{\sqrt{x}} dx\]

\[\int \left( 3x + 4 \right)^2 dx\]

\[\int\left( \sec^2  x + {cosec}^2  x \right)  dx\]

\[\int \left( \tan x + \cot x \right)^2 dx\]

\[\int\frac{\left( x^3 + 8 \right)\left( x - 1 \right)}{x^2 - 2x + 4} dx\]

If f' (x) = x − \[\frac{1}{x^2}\]  and  f (1)  \[\frac{1}{2},    find  f(x)\]

 


\[\int\frac{2x + 1}{\sqrt{3x + 2}} dx\]

\[\int\left( 5x + 3 \right) \sqrt{2x - 1} dx\]

\[\int\sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} dx\]

\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x e^x \right)} dx\]

\[\int\frac{\sin \left( \text{log x} \right)}{x} dx\]

\[\  ∫    x   \text{ e}^{x^2} dx\]

` ∫  tan^3    x   sec^2  x   dx  `

\[\int {cosec}^4  \text{ 3x } \text{ dx } \]

Evaluate the following integrals:

\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]

\[\int\frac{1}{\sqrt{\left( 2 - x \right)^2 - 1}} dx\]

\[\int\frac{1}{1 + x - x^2}  \text{ dx }\]

\[\int\frac{1}{\sqrt{7 - 6x - x^2}} dx\]

\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]

\[\int\frac{2x + 5}{x^2 - x - 2} \text{ dx }\]

\[\int\frac{x^2}{x^2 + 6x + 12} \text{ dx }\]

\[\int\frac{6x - 5}{\sqrt{3 x^2 - 5x + 1}} \text{ dx }\]

\[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\]

\[\int\left( x + 1 \right) \text{ e}^x \text{ log } \left( x e^x \right) dx\]

\[\int\frac{x^2 + 6x - 8}{x^3 - 4x} dx\]

\[\int\frac{1}{1 + x + x^2 + x^3} dx\]

Evaluate the following integral:

\[\int\frac{x^2}{1 - x^4}dx\]

\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{  dx }\]

\[\int\frac{\left( x - 1 \right)^2}{x^4 + x^2 + 1} \text{ dx}\]

\[\int\frac{1}{\left( x^2 + 1 \right) \sqrt{x}} \text{ dx }\]

\[\int\frac{1}{\left( 2 x^2 + 3 \right) \sqrt{x^2 - 4}} \text{ dx }\]

If \[\int\frac{\sin^8 x - \cos^8 x}{1 - 2 \sin^2 x \cos^2 x} dx\]


The primitive of the function \[f\left( x \right) = \left( 1 - \frac{1}{x^2} \right) a^{x + \frac{1}{x}} , a > 0\text{ is}\]


\[\int\frac{x^3}{x + 1}dx\] is equal to

\[\int\frac{x^2}{\left( x - 1 \right)^3} dx\]

\[\int\sqrt{\frac{1 - x}{x}} \text{ dx}\]


\[\int\frac{1}{\sin x \left( 2 + 3 \cos x \right)} \text{ dx }\]

\[\int\frac{\sin^5 x}{\cos^4 x} \text{ dx }\]

\[\int\sqrt{a^2 + x^2} \text{ dx }\]

\[\int x\sqrt{\frac{1 - x}{1 + x}} \text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×