Grade 12Integral Calculus

Integrate the following:-
∫(sin^6 x+cos^6 x)/sin^2 x.cos^2 x dx

Sudipan Datta

8 Years agoGrade 12

3 Answers

Arun

8 Years ago

First note that sin6(x)+cos6(x) is a sum of two cubes (sin2(x))3+(cos2(x))3 so it can be factored using a3+b3=(a+b)(a2−ab+b2) and the Pythagorean identity to get

sin6(x)+cos6(x)=(sin2(x))3+(cos2(x))3

=(sin2(x)+cos2(x))(sin4(x)−sin2(x)cos2(x)+cos4(x))

=sin4(x)−sin2(x)cos2(x)+cos4(x).

Therefore,

sin6x+cos6xsin2(x)cos2(x)=sin4(x)−sin2(x)cos2(x)+cos4(x)sin2(x)cos2(x)

=tan2(x)−1+cot2(x)

But tan2(x)=sec2(x)−1 and cot2(x)=csc2(x)−1. Hence,

sin6x+cos6xsin2(x)cos2(x)=sec2(x)+csc2(x)−3 and thus

∫sin6(x)+cos6(x)sin2(x)cos2(x) dx=∫sec2(x)+csc2(x)−3 dx

=tan(x)−cot(x)−3x+C

tuhar

7 Years ago

=sin6(x)/sin²x.cos²x+cos6(x)/sin²x.cos²x

=sin⁴x/cos²x+cos⁴x/sin²x

=(1-cos²x)²/cos²x + (1-sin²x)^2/sin²x

=(1-2cos²x+cos⁴x)/cos²x + (1-2sin²x+sin⁴x)/sin²x

=sec²x -2 + cos²x + cosec²x -2 + sin²x

=sec²x + cosec²x -4 + 1

=sec²x + cosec²x -3

so, by integration we get,

=tan x – cot x -3x + C

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

first resolve , sin^6x + cos^6x
= (sin²x)³ + (cos²x)³
= (sin²x + cos²x)(sin⁴x + cos⁴x - sin²x.cos²x)
=(sin²x + cos²x) {(sin²x + cos²x)² - 3sin²x.cos²x}
but we know, sin²x + cos²x = 1
so, sin^6x + cos^6x = (1 - 3sin²x.cos²x)
or, you can write cos^6x + cos^6x = sin²x + cos²x - 3sin²x.cos²x [ as we know, sin²x + cos²x = 1 ]