Question

Please,help me!It is integral with substitution

Answer 1

Question :

Integrate

$1) \int \dfrac{cosx \: dx}{4 + sin {}^{2}x}$

$2) \int \dfrac{dx}{x(1 + ln {}^{2} x)}$

Theory:

Integration is the inverse process of Differentiation.

•Methods of integration

Integration by substitution :

The given integral $\int f(x)dx$ can be transformed into another by changing the independent variable x to t by substituting x = g(t)

•Formula's used :

$1) \int \frac{1}{x {}^{2} + a {}^{2} } = \frac{1}{a} \tan {}^{ - 1} ( \frac{x}{a} )$

$2) \int \frac{dx}{1 + x {}^{2} } = \tan {}^{ - 1} (x)$

Solution :

$1) I= \int \bf \dfrac{cosx \: dx}{4 + sin {}^{2} x}$

Let sinx = t

$\frac{dy}{dx} = cosx\implies \frac{dt}{cosx} = dx$

$I= \int \frac{ \cancel cosx}{4 + t {}^{2} } \times \frac{dt}{ \cancel cosx}$

$I= \int \frac{dt}{4+ t {}^{2} }$

$I= \int \frac{dt}{2 {}^{2} + t {}^{2} }$

use formula : $\int\:\frac{dx}{x{}^{2}+a{}^{2}}$

$I = \frac{1}{2} \tan {}^{ - 1} ( \frac{t}{2} ) + c$

Put t = sinx

$I= \dfrac{1}{2} \tan( \frac{ \sin x}{2} ) + c$

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$2) I=\int \bf \dfrac{dx}{x(1 +ln {}^{2} x) }$

Let lnx = t ⇒xdt = dt

$\int \bf \frac{dx}{x(1 + ln {}^{2}x) } = \int \frac{ \cancel xdt}{ \cancel x(1 + t {}^{2} )}$

$I = \int \frac{dt}{1 + t {}^{2} }$

$I= \tan {}^{ - 1} (t) + c$

put t = ln x

$I= \tan {}^{ - 1} (lnx) + c$