Evaluate: ∫ 2 Cos 2 X − Cos 2 X Cos 2 X D X - Mathematics

प्रश्न

Evaluate:

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

योग

उत्तर

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

shaalaa.com

Evaluation of Definite Integrals by Substitution

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 5.2 Q 5.1 Q 6

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

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12

अध्याय 19 Indefinite Integrals
Exercise 19.01 | Q 5.2 | पृष्ठ ४

वीडियो ट्यूटोरियलVIEW ALL [1]

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

video tutorial00:18:12

संबंधित प्रश्न

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

Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`

Evaluate : `int1/(3+5cosx)dx`

Evaluate :

`∫_(-pi)^pi (cos ax−sin bx)^2 dx`

find `∫_2^4 x/(x^2 + 1)dx`

Evaluate :

`∫_0^π(4x sin x)/(1+cos^2 x) dx`

If `int_0^a1/(4+x^2)dx=pi/8` , find the value of a.

Evaluate the integral by using substitution.

`int_0^1 x/(x^2 +1)`dx

Evaluate the integral by using substitution.

`int_0^2 dx/(x + 4 - x^2)`

Evaluate the integral by using substitution.

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

Evaluate of the following integral:

(i) \[\int x^4 dx\]

Evaluate of the following integral:

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

Evaluate:

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

Evaluate:

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

Evaluate the following definite integral:

\[\int_0^1 \frac{1}{\sqrt{\left( x - 1 \right)\left( 2 - x \right)}}dx\]