-1

Sin^2x cos^6x I tried to let sin^2x = u , cos^6x = v and find du ,dv but couldn't solve it . Can someone help me to learn how to start

prog
  • 19
  • Welcome to Math SX! For me, your problem is perfectly clear. Just a hint: linearise with the basic trigonometry formula: $$\sin a\cos b=\dfrac12\bigl(\sin (a+b)+\sin(a-b)\bigr).$$ – Bernard Nov 15 '17 at 09:55

1 Answers1

1

Hint:

For $$\int\sin^{2m}x\cos^{2n}x\ dx$$

we need to use $$\cos2y=1-2\sin^2y=2\cos^2y-1$$

OR How to prove Euler's formula: $e^{it}=\cos t +i\sin t$?,

$$2^{2m+2n}(-1)^m\sin^{2m}x\cos^{2n}x=(2i\sin x)^{2m}(2\cos x)^{2n}$$ $$=(e^{ix}-e^{ix})^{2m}(e^{ix}+e^{ix})^{2n}$$

to find terms of the form $\int\cos2rx\ dx$

lab bhattacharjee
  • 274,582