मराठी

∫ 1 1 − Sin X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\frac{1}{1 - \sin x} dx\]
बेरीज

उत्तर

\[\int\frac{dx}{1 - \sin x}\]
\[ = \int\frac{\left( 1 + \sin x \right)}{\left( 1 - \sin x \right) \times \left( 1 + \sin x \right)}dx\]
\[ = \int\left( \frac{1 + \sin x}{1 - \sin^2 x} \right)dx\]
\[ = \int\left( \frac{1 + \sin x}{\cos^2 x} \right)dx\]
\[ = \int\left( \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} \times \frac{1}{\cos x} \right)dx\]
\[ = \int\left( \sec^2 x + \sec x \tan x \right)dx\]
\[ = \tan  x + \sec x + C\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.02 | Q 30 | पृष्ठ १५

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

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

` ∫  {cosec x} / {"cosec x "- cot x} ` dx      


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


If f' (x) = 8x3 − 2xf(2) = 8, find f(x)


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

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


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

\[\int\frac{a}{b + c e^x} dx\]

\[\int\left\{ 1 + \tan x \tan \left( x + \theta \right) \right\} dx\]

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

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

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

\[\int\sqrt {e^x- 1}  \text{dx}\] 

\[\int\frac{1}{x^2 - 10x + 34} dx\]

\[\int\frac{1}{\sqrt{\left( x - \alpha \right)\left( \beta - x \right)}} dx, \left( \beta > \alpha \right)\]

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

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

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

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

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

\[\int\frac{1}{5 + 7 \cos x + \sin x} dx\]

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

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

` ∫    sin x log  (\text{ cos x ) } dx  `

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

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

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

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

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

Evaluate the following integral:

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

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


\[\int\frac{\sin^6 x}{\cos^8 x} dx =\]

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


\[\int\frac{\sin^6 x}{\cos x} \text{ dx }\]

\[\int \tan^3 x\ \sec^4 x\ dx\]

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

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

\[\int\frac{x^2}{x^2 + 7x + 10}\text{ dx }\]

Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×