मराठी

∫ − Sin X + 2 Cos X 2 Sin X + Cos X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
बेरीज
Advertisements

उत्तर

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

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 29 | पृष्ठ ४८

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

 


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

\[\int\frac{\tan x}{\sec x + \tan x} dx\]

\[\int\frac{1}{\sqrt{x + a} + \sqrt{x + b}} dx\]

\[\int \left( e^x + 1 \right)^2 e^x dx\]

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

\[\int\frac{3x + 5}{\sqrt{7x + 9}} dx\]

`∫     cos ^4  2x   dx `


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

` ∫   cos  3x   cos  4x` dx  

\[\int\frac{1}{x (3 + \log x)} dx\]

\[\int \cot^n {cosec}^2 \text{ x dx } , n \neq - 1\]

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

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

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

\[\int\frac{1}{\sqrt{8 + 3x - x^2}} dx\]

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

\[\int\frac{1}{\sqrt{\left( 1 - x^2 \right)\left\{ 9 + \left( \sin^{- 1} x \right)^2 \right\}}} dx\]

\[\int\frac{\cos x}{\cos 3x} \text{ dx }\]

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

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

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

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

\[\int e^x \left( \tan x - \log \cos x \right) dx\]

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

\[\int\sqrt{3 - 2x - 2 x^2} \text{ dx}\]

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

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

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

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

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

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

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

\[\int\frac{x^3}{\sqrt{x^8 + 4}} \text{ dx }\]


\[\int \sec^6 x\ dx\]

\[\int\sqrt{a^2 - x^2}\text{  dx }\]

\[\int \sec^{- 1} \sqrt{x}\ dx\]

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×