Skip to main contentSkip to solution

If ∫sin⁡6xcos⁡10x dx=−2atan⁡9x+btan⁡7x+c\int \frac{\sin^6 x}{\cos^{10} x}\,dx = -2a\tan^9 x + b\tan^7 x + c, (where c is an arbitrary constant), then which one of the following is TRUE?

Solution

✅ Correct Option: 2

Write the integrand as tan⁡6xsec⁡4x=tan⁡6x(1+tan⁡2x)sec⁡2x\tan^6x\sec^4x=\tan^6x(1+\tan^2x)\sec^2x. With t=tan⁡xt=\tan x, the integral is t77+t99\frac{t^7}{7}+\frac{t^9}{9}. Comparing, −2a=19-2a=\frac19 and b=17b=\frac17, so 18a=−118a=-1 and 7b=17b=1, giving 18a+7b=018a+7b=0.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question