If ∫ co s ^ 4 xdx si n ^ 3 x (si n ^ 5 x + co s ^ 5 x) ^ (3)/(5) = - (1)/(K) (1 + co t ^ P x) ^ (K)/(P) + C , then the value of K + P is equal to (where C is the constant of integration)

Correct answer: 7 Let, I = ∫ co s ^ 4 xdx si n ^ 6 x(1 + c ot ^ 5 x)(3)/(5) = ∫ se c ^ 2 xdx ta n ^ 6 x(1 + ((1)/( x)) ^ 5 )(3)/(5) Substituting, 1 + 1 (tan x) ^ 5 = t and - 5 ( ^ 6 x) ( ^ 2 x)dx = dt I = - (1)/(5) ∫ dt t ^ 3 / 5 = ((- 1)/(5))(t(2)/(5))/(((2)/(5))) + C = - (1)/(2)∫ (1 + co t ^ 5 x) ^ (2)/(5) + C Hence, K = 2, P = 5

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If ∫ co s ^ 4 xdx si n ^ 3 x (si n ^ 5 x + co s ^ 5 x) ^ (3)/(5) =...

∫ \dfrac{co{s}^{4}xdx}{si{n}^{3}x{\left(si{n}^{5}x + co{s}^{5}x\right)}^{\dfrac{3}{5}}} = - \dfrac{1}{K}{\left(1 + co{t}^{P}x\right)}^{\dfrac{K}{P}} + C

, then the value of

K + P

is equal to (where

C

is the constant of integration)

Correct answer

Step-by-step solution

Let,

I = ∫ \dfrac{co{s}^{4}xdx}{si{n}^{6}x\left(1 + c{ot}^{5}x\right)\dfrac{3}{5}} = ∫ \dfrac{se{c}^{2}xdx}{ta{n}^{6}x\left(1 + {\left(\dfrac{1}{\tan x}\right)}^{5}\right)\dfrac{3}{5}}

Substituting,

1 + \dfrac{1}{{\left(tan x\right)}^{5}} = t

and

- \dfrac{5}{\left(\tan^{6} x\right)}\left(\sec^{2} x\right)dx = dt

I = - \dfrac{1}{5} ∫ \dfrac{dt}{{t}^{3 / 5}} = \left(\dfrac{- 1}{5}\right)\dfrac{t\dfrac{2}{5}}{\left(\dfrac{2}{5}\right)} + C

= - \dfrac{1}{2}∫ {\left(1 + co{t}^{5}x\right)}^{\dfrac{2}{5}} + C

Hence,

K = 2, P = 5

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