Mathematics

# Evaluate the following integral$\int { \cfrac { \cos { 2x } }{ { \left( \cos { x } +\sin { x } \right) }^{ 2 } } } dx\quad$

##### SOLUTION
$I=\displaystyle\int{\dfrac{\cos{2x}}{{\left(\cos{x}+\sin{x}\right)}^{2}}dx}$

$I=\displaystyle\int{\dfrac{{\cos}^{2}{x}-{\sin}^{2}{x}}{{\left(\cos{x}+\sin{x}\right)}^{2}}dx}$

$I=\displaystyle\int{\dfrac{\left(\cos{x}+\sin{x}\right)\left(\cos{x}-\sin{x}\right)}{{\left(\cos{x}+\sin{x}\right)}^{2}}dx}$

$I=\displaystyle\int{\dfrac{\left(\cos{x}-\sin{x}\right)}{\left(\cos{x}+\sin{x}\right)}dx}$

Let $t=\cos{x}+\sin{x}\Rightarrow\,dt=\left(-\sin{x}+\cos{x}\right)dx$

$I=\displaystyle\int{\dfrac{dt}{t}}$

$I=\log{\left|t\right|}+c$

$\therefore\,I=\log{\left|\cos{x}+\sin{x}\right|}+c$ where $t=\cos{x}+\sin{x}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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Q5 Passage Hard
In calculating a number of integrals we had to use the method of integration by parts several times in succession.
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$\displaystyle \int u\left ( x \right )v\left ( x \right )dx=u\left ( x \right )v_{1}-u'\left ( x \right )v_{2}\left ( x \right )+u''\left ( x \right )v_{3}\left ( x \right )+...+\left ( -1 \right )^{n-1}u^{n-1}\left ( x \right )V_{n}\left ( x \right ) \\ -\left ( -1 \right )^{n-1}\int u^{n}\left ( x \right )V_{n}\left ( x \right )dx$
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Of course, we assume that all derivatives and integrals appearing in this formula exist. The use of the generalized formula for integration by parts is especially useful when  calculating $\displaystyle \int P_{n}\left ( x \right )Q\left ( x \right )dx,$ where $\displaystyle P_{n}\left ( x \right )$ is polynomial of degree n and the factor Q(x) is such that it can be integrated successively n+1 times.