हिंदी

∫ ( X + 1 ) E X Cos 2 ( X E X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[\int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x \cdot e^x \right)} dx\]
\[\text{Let x e}^x = t\]
\[ \Rightarrow \left( 1 \cdot e^x + \text{x  e}^x \right) = \frac{dt}{dx}\]
\[ \Rightarrow \left( x + 1 \right) e^x dx = dt\]
\[Now, \int\frac{\left( x + 1 \right) e^x}{\cos^2 \left( x \cdot e^x \right)} dx\]
\[ = \int\frac{dt}{\cos^2 t}\]
\[ = \int \sec^2 \text{t dt}\]
\[ = \tan \left( t \right) + C\]
` = tan   ( x  e^x)   + C `

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Indefinite Integrals - Exercise 19.09 [पृष्ठ ५८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.09 | Q 37 | पृष्ठ ५८

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

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

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

If f' (x) = x + bf(1) = 5, f(2) = 13, find f(x)


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

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

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

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

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

\[\int\sqrt {e^x- 1}  \text{dx}\] 

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

\[\int\frac{x}{3 x^4 - 18 x^2 + 11} dx\]

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

\[\int\frac{1}{x\sqrt{4 - 9 \left( \log x \right)^2}} dx\]

\[\int\frac{\left( 3 \sin x - 2 \right) \cos x}{5 - \cos^2 x - 4 \sin x} dx\]

\[\int\frac{x + 2}{2 x^2 + 6x + 5}\text{  dx }\]

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

\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{  dx }\]

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

\[\int\frac{1}{4 \cos x - 1} \text{ dx }\]

\[\int\frac{1}{4 + 3 \tan x} dx\]

\[\int2 x^3 e^{x^2} dx\]

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

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

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

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

\[\int\frac{5x}{\left( x + 1 \right) \left( x^2 - 4 \right)} dx\]

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

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

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

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

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

If \[\int\frac{1}{5 + 4 \sin x} dx = A \tan^{- 1} \left( B \tan\frac{x}{2} + \frac{4}{3} \right) + C,\] then


If \[\int\frac{\cos 8x + 1}{\tan 2x - \cot 2x} dx\]


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


\[\int\sqrt{\frac{x}{1 - x}} dx\]  is equal to


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


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

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

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

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×