Question

please solve this as fast as possible​

Answer 1

Answer:

see step by step.....

Step-by-step explanation:

∫

sin

2

x.cos

2

x

sin

6

x+cos

6

x

dx

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 ]

now, \int{\frac{(sin^2x+cos^2x-3sin^2x.cos^2x)}{sin^2x.cos^2x}}\,dx∫

sin

2

x.cos

2

x

(sin

2

x+cos

2

x−3sin

2

x.cos

2

x)

dx

= \int{sec^2x+cosec^2x-3}\,dx∫sec

2

x+cosec

2

x−3dx

= \int{sec^2x}\,dx-\int{cosec^2x}\,dx-\int{3}\,dx∫sec

2

xdx−∫cosec

2

xdx−∫3dx

= tanx - cotx - 3x + C

hence, \int{\frac{sin^6x+cos^6x}{sin^2x.cos^2x}}\,dx∫

sin

2

x.cos

2

x

sin

6

x+cos

6

x

dx

= tanx - cotx - 3x + C