Evaluate: ∫ √ 1 + Cos 2 X 2 D X - Mathematics

प्रश्न

Evaluate:

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

बेरीज

उत्तर

\[\int\sqrt{\frac{1 + \cos 2x}{2}}dx\]
\[ \int\sqrt{\frac{\text{2 cos}^2 x}{2}}dx \left[ \therefore 1 + \cos2A = 2 \cos^2 A \right]\]
\[ = \int\ \text{cos x dx}\]
\[ = \sin x + C\]

shaalaa.com

Evaluation of Definite Integrals by Substitution

या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?

Q 2.1 Q 1.8 Q 2.2

पाठ 19: Indefinite Integrals - Exercise 19.01 [पृष्ठ ४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12

पाठ 19 Indefinite Integrals
Exercise 19.01 | Q 2.1 | पृष्ठ ४

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

view
व्हिडिओ ट्यूटोरियल For All Subjects
Evaluation of Definite Integrals by Substitution

video tutorial00:18:12

संबंधित प्रश्‍न

Evaluate: `int1/(xlogxlog(logx))dx`

Evaluate `∫_0^(3/2)|x cosπx|dx`

Evaluate `int_(-1)^2|x^3-x|dx`

Evaluate :

`∫_0^π(4x sin x)/(1+cos^2 x) dx`

Evaluate: `intsinsqrtx/sqrtxdx`

Evaluate the integral by using substitution.

`int_0^2 xsqrt(x+2)` (Put x + 2 = `t^2`)

Evaluate the integral by using substitution.

`int_(-1)^1 dx/(x^2 + 2x + 5)`

`int 1/(1 + cos x)` dx = _____

A) `tan(x/2) + c`

B) `2 tan (x/2) + c`

C) -`cot (x/2) + c`

D) -2 `cot (x/2)` + c

Evaluate of the following integral:

\[\int\frac{1}{x^5}dx\]

Evaluate of the following integral:

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

Evaluate of the following integral:

\[\int 3^{2 \log_3} {}^x dx\]

Evaluate:

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

Evaluate :

\[\int\frac{e^{6 \log_e x} - e^{5 \log_e x}}{e^{4 \log_e x} - e^{3 \log_e x}}dx\]