मराठी

∫ X Sin X ( Sin X X + Cos X . Log X ) D X I S E Q U a L T O (A) Xsin X + C (B) Xsin X Cos X + C (C) ( X Sin X ) 2 2 + C (D) None of These - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int x^{\sin x} \left( \frac{\sin x}{x} + \cos x . \log x \right) dx\] is equal to

पर्याय

  •  xsin x + C

  •  xsin x cos x + C

  • \[\frac{\left( x^{\sin x} \right)^2}{2} + C\]

  • none of these

MCQ

उत्तर

 xsin x + 

\[\text{ Let I } = \int x^{\sin x} \left( \frac{\sin x}{x} + \cos x \cdot \log x \right)dx\]
\[\text{ Putting  x}^{\sin x} = t\]
\[ \Rightarrow \ln \left( x \right)^{\sin x} = \ln t\]
\[ \Rightarrow \sin x \cdot \ln x = \ln t\]
\[ \Rightarrow \left( \sin x \times \frac{1}{x} + \cos x \ln x \right)dx = \frac{1}{t}dt\]
\[ \therefore I = \int t \cdot \frac{dt}{t}\]
\[ = t + C\]
\[ = x^{\sin x} + C .............\left( \because t = x^{\sin x} \right)\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - MCQ [पृष्ठ २००]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
MCQ | Q 5 | पृष्ठ २००

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

\[\int\left\{ \sqrt{x}\left( a x^2 + bx + c \right) \right\} dx\]

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

\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]

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

\[\int \tan^2 \left( 2x - 3 \right) dx\]


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


` ∫  {sec  x   "cosec " x}/{log  ( tan x) }`  dx


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

\[\int\frac{2 \cos 2x + \sec^2 x}{\sin 2x + \tan x - 5} dx\]

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

` ∫  sec^6   x  tan    x   dx `

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

` ∫  {1}/{a^2 x^2- b^2}dx`

\[\int\frac{\cos x}{\sin^2 x + 4 \sin x + 5} dx\]

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

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

\[\int \log_{10} x\ dx\]

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

\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]

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

\[\int x \sin^3 x\ dx\]

\[\int x \cos^3 x\ dx\]

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

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

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

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

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

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

\[\int\frac{5}{\left( x^2 + 1 \right) \left( x + 2 \right)} dx\]

\[\int\frac{x^2 + 9}{x^4 + 81} \text{ dx }\]

 


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

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

\[\int\frac{1}{1 + \tan x} dx =\]

The value of \[\int\frac{\sin x + \cos x}{\sqrt{1 - \sin 2x}} dx\] is equal to


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

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

\[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\]

\[\int\frac{5x + 7}{\sqrt{\left( x - 5 \right) \left( x - 4 \right)}} \text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×