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int(cos xdx)/(sqrt(sin^(2)x+4sin x-1))

∫cos⁡xdxsin⁡2x+4sin⁡x−1 \int \frac{\cos x d x}{\sqrt{\sin ^{2} x+4 \sin x-1}}

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Find indefinite integrals using the substitution and by partsright chevron icon

Full solution

Q. ∫cos⁡xdxsin⁡2x+4sin⁡x−1 \int \frac{\cos x d x}{\sqrt{\sin ^{2} x+4 \sin x-1}}
  1. Identify Integral: Identify the integral to be solved. I=∫cos⁡xsin⁡2(x)+4sin⁡x−1 dxI = \int \frac{\cos x}{\sqrt{\sin^2(x) + 4\sin x - 1}} \, dx
  2. Simplify Expression: Simplify the expression inside the square root if possible.\newlineThe expression under the square root is a quadratic in terms of sin⁡(x)\sin(x), which can be factored or completed to a square if it represents a perfect square trinomial.
  3. Factor Quadratic: Attempt to factor the quadratic expression sin⁡2(x)+4sin⁡x−1\sin^2(x) + 4\sin x - 1. The quadratic does not factor easily into real factors, so we consider completing the square to transform it into a form that might be more easily integrated. sin⁡2(x)+4sin⁡x−1=(sin⁡(x)+2)2−5\sin^2(x) + 4\sin x - 1 = (\sin(x) + 2)^2 - 5
  4. Rewrite Integral: Rewrite the integral with the completed square. \newlineI=∫cos⁡x(sin⁡(x)+2)2−5 dxI = \int \frac{\cos x}{\sqrt{(\sin(x) + 2)^2 - 5}} \, dx
  5. Look for Substitution: Look for a substitution that will simplify the integral. Let u=sin⁡(x)+2u = \sin(x) + 2, then du=cos⁡(x) dxdu = \cos(x) \, dx. This substitution will simplify the integral.
  6. Perform Substitution: Perform the substitution.\newlineI=∫1u2−5 duI = \int \frac{1}{\sqrt{u^2 - 5}} \, du
  7. Recognize Standard Form: Recognize the integral as a standard form. The integral ∫1u2−a2 du\int \frac{1}{\sqrt{u^2 - a^2}} \, du is a standard form that corresponds to the inverse hyperbolic function arcsinh(ua)\text{arcsinh}(\frac{u}{a}) or the logarithmic form ln⁡∣u+u2−a2∣\ln|u + \sqrt{u^2 - a^2}|.
  8. Integrate Using Form: Integrate using the standard form. I=ln⁡∣u+u2−5∣+CI = \ln|u + \sqrt{u^2 - 5}| + C, where CC is the constant of integration.
  9. Substitute Back: Substitute back to the original variable.\newlineI=ln⁡∣sin⁡(x)+2+(sin⁡(x)+2)2−5∣+CI = \ln|\sin(x) + 2 + \sqrt{(\sin(x) + 2)^2 - 5}| + C
  10. Simplify Final Answer: Simplify the final answer if possible.\newlineI=ln⁡∣sin⁡(x)+2+sin⁡2(x)+4sin⁡(x)−1∣+CI = \ln|\sin(x) + 2 + \sqrt{\sin^2(x) + 4\sin(x) - 1}| + C

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