मराठी

∫ Sin 2 X a 2 + B 2 Sin 2 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{\sin 2x}{a^2 + b^2 \sin^2 x} dx\]
बेरीज
Advertisements

उत्तर

\[\text{Let I} = \int\frac{\sin 2x}{a^2 + b^2 \sin^2 x}dx\]
\[\text{Putting}\ \sin^2 x = t\]
\[ \Rightarrow 2\sin x . \cos x = \frac{dt}{dx}\]
\[ \Rightarrow \sin 2x = \frac{dt}{dx}\]
\[ \Rightarrow \text{sin 2x dx} = dt\]
\[ \therefore I = \int\frac{1}{a^2 + b^2 t}dt\]
\[ = \frac{1}{b^2} \text{ln }\left| a^2 + b^2 t \right| + C\]
\[ = \frac{1}{b^2} \text{ln }\left| a^2 + b^2 \sin^2 x \right| + C \left[ \because t = \sin^2 x \right]\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.08 [पृष्ठ ४८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.08 | Q 38 | पृष्ठ ४८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

 
\[\int\frac{\cos x}{1 - \cos x} \text{dx }or \int\frac{\cot x}{\text{cosec         } {x }- \cot x} dx\]

\[\int \cos^2 \text{nx dx}\]

Integrate the following integrals:

\[\int\text{sin 2x  sin 4x    sin 6x  dx} \]

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

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

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

\[\int\frac{\sec^2 \sqrt{x}}{\sqrt{x}} dx\]

\[\int \tan^3 \text{2x sec 2x dx}\]

\[\int\frac{1}{\sin x \cos^3 x} dx\]

\[\int\frac{\cos x}{\sqrt{4 - \sin^2 x}} dx\]

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

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

\[\int\frac{1}{\sin x + \sqrt{3} \cos x} \text{ dx  }\]

\[\int\frac{1}{p + q \tan x} \text{ dx  }\]

\[\int x^2 \text{ cos x dx }\]

\[\int x^2 \sin^2 x\ dx\]

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

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

\[\int \sin^3 \sqrt{x}\ dx\]

\[\int e^x \left( \cot x - {cosec}^2 x \right) dx\]

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

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

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

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

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

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

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

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

\[\int\frac{x^2 + 1}{x^4 + 7 x^2 + 1} 2 \text{ dx }\]

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

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

Write a value of

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

` \int \text{ x} \text{ sec x}^2 \text{  dx  is  equal  to }`

 


\[\int\frac{\left( 2^x + 3^x \right)^2}{6^x} \text{ dx }\] 

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

\[\int\frac{\sqrt{1 - \sin x}}{1 + \cos x} e^{- x/2} \text{ dx}\]

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

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×