3

$\int\ e^{-x/2}\sqrt(sin-1)/(cosx+1)$

The result that I'm getting here contains a cosecx term but the answer has secx term! Please help.

PalmTree
  • 41
  • 3
    Do you mind showing your work? It will help us see what you have tried and where you lacked. – prog_SAHIL Aug 19 '18 at 11:08
  • You say you completed the square ... what about looking for a suitable substitution once you have it in the form $\int \sqrt(2(x+\frac{3}{4})^2+\frac{23}{8}dx$ – Mandelbrot Aug 19 '18 at 11:14
  • 4
    See https://math.stackexchange.com/questions/390080/definite-integral-of-square-root-of-polynomial – Robert Z Aug 19 '18 at 11:15
  • Did you just edit the question ? If you have another doubt, kindly ask it as a new question. – prog_SAHIL Aug 20 '18 at 14:39

3 Answers3

1

Do you know the integral?

$$\int{\sqrt{x^2+a^2}}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\ln(\frac{{x+\sqrt{x^2+a^2}}}{a})+C$$

prog_SAHIL
  • 2,543
  • 13
  • 27
0

Hint: Substitute $z=(x+\frac{3}{4})$ after completing the square.

tarit goswami
  • 3,009
  • 1
  • 12
  • 27
0

Hint: Use the so-called Euler Substitution: $$\sqrt{2x^2+3x+4}=x\sqrt{2}+t$$ then we get $$x=\frac{t^2-4}{3-2t\sqrt{2}}$$ $$dx=\frac{-2\sqrt{2}t^2+6t-8\sqrt{2}}{(3-2t\sqrt{2})^2}dt$$ and $$\sqrt{2x^2+3x+4}=\frac{-\sqrt{2}t^2+3t-4\sqrt{2}}{3-2t\sqrt{2}}$$

Dr. Sonnhard Graubner
  • 95,283