Question

integration of Cos^5X

Answer 1

=∫(cos5x)dx

From trigonometric identity, which is

cos2x+sin2x=1, ⇒cos2x=1−sin2x

=∫(cos4x)⋅cos(x)dx

=∫(cos2x)2⋅cos(x)dx

=∫(1−sin2x)2⋅cos(x)dx .. (i)

let's assume sinx=t, ⇒(cosx)dx=dt

substituting this in the (i), we get

=∫(1−t2)2dt

Now using expansion of (1−y)2=1+y2−2y, yields,

=∫(1+t4−2t2)dt

=∫dt+∫t4dt−2∫t2dt

=t+t55−2⋅t33+c, where c is a constant

=t+t55−23⋅t3+c, where c is a constant

now substituting t back gives,

=sinx+(sinx)55−23⋅(sinx)3+c, where cis a constant

=sinx+sin5x5−23⋅sin3x+c, where c is

Answer 2

Questions:-

$\displaystyle\int \rm \: { \cos}^{5} x \: dx$

Solution:-

There is no direct formulae to find this integral. so, let's simplify the given trigonometric term

cos⁵x = cos⁴x.cosx = (cos²x)².cosx

[We know cos²x = 1- sin²x ]

so put it  we get,

= (1-sin²x)².cosx

Now we can find the integration

so put it,

$= \displaystyle \int \rm \: {(1 - { \sin}^{2} x})^{2} . \cos x \: dx$

put sin x = t

cos x dx = dt

$= \displaystyle \int \rm \: {(1 - { t}^{2}})^{2} . dt$

$= \displaystyle \int \rm \: ( 1 + {t}^{4} - 2 {t}^{2} )dt$

$= \displaystyle \int \rm {t}^{4} dt + \int \: 1dt \: - \int \: 2 {t}^{2} dt$

$\rm = \dfrac{ {t}^{5} }{5} + t - 2 \times \dfrac{ {t}^{3} }{3} + c$

Now put the value of  t we get

$\rm \: = \dfrac{ { \sin }^{5}x }{5} + \sin x - \dfrac{ { 2\sin }^{3} x}{3} + c$

$\rm = \sin x - \dfrac{2}{3} { \sin}^{3} x + \dfrac{1}{5} { \sin}^{5} x + c$

Some Important integration formulae

$\displaystyle\int \rm{x}^{n }dx = \dfrac{ {x}^{n + 1} }{n + 1}$

$\displaystyle\int \rm{ \sin x \: dx = - \cos x + c}$

$\displaystyle\int \rm{ \ \cos x \: dx = \: \sin x + c}$

$\displaystyle\int \rm{ { \sec }^{2} x \: dx = \tan x + c}$

$\displaystyle\int \rm{ { \csc }^{2} x \: dx = - \cot x + c}$

$\displaystyle \int \rm \sec x . \tan x = \sec x + c$

$\displaystyle \int \rm \csc x . \cot x = - \csc x + c$

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