We read your photo as:

Q13. Compute the integral: integral(sin2 x / (1 + cos x)) dx

Mathematics

ScanSolve AI Answer

Step-by-step solution

20 steps

Q10. Evaluate the integral: ∫(xx+1)dx\int \left(\frac{x}{x+1}\right) dx

  1. Rewrite the integrand using algebraic manipulation.

    xx+1=x+1−1x+1=1−1x+1\frac{x}{x+1} = \frac{x+1-1}{x+1} = 1 - \frac{1}{x+1}

  2. Integrate each term.

    ∫(1−1x+1)dx=∫1 dx−∫1x+1 dx\int \left(1 - \frac{1}{x+1}\right) dx = \int 1 \, dx - \int \frac{1}{x+1} \, dx

  3. Perform the integration.

    ∫1 dx=x\int 1 \, dx = x ∫1x+1 dx=ln⁡∣x+1∣\int \frac{1}{x+1} \, dx = \ln|x+1|

  4. Combine the results and add the constant of integration.

    x−ln⁡∣x+1∣+Cx - \ln|x+1| + C

    The final answer is x−ln⁡∣x+1∣+C\boxed{x - \ln|x+1| + C}.

    Q11. Evaluate the integral: ∫tan⁡2x dx\int \tan^2 x \, dx

  5. Use the trigonometric identity tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1.

    ∫tan⁡2x dx=∫(sec⁡2x−1) dx\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx

  6. Integrate each term.

    ∫sec⁡2x dx−∫1 dx\int \sec^2 x \, dx - \int 1 \, dx

  7. Perform the integration.

    ∫sec⁡2x dx=tan⁡x\int \sec^2 x \, dx = \tan x ∫1 dx=x\int 1 \, dx = x

  8. Combine the results and add the constant of integration.

    tan⁡x−x+C\tan x - x + C

    The final answer is tan⁡x−x+C\boxed{\tan x - x + C}.

    Q12. Compute the indefinite integral: ∫1−sin⁡2x dx\int \sqrt{1-\sin 2x} \, dx

  9. Use the identities 1=sin⁡2x+cos⁡2x1 = \sin^2 x + \cos^2 x and sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2 \sin x \cos x to rewrite the expression under the square root.

    1−sin⁡2x=sin⁡2x+cos⁡2x−2sin⁡xcos⁡x=(cos⁡x−sin⁡x)21 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2

  10. Substitute this back into the integral.

    ∫(cos⁡x−sin⁡x)2 dx=∫∣cos⁡x−sin⁡x∣ dx\int \sqrt{(\cos x - \sin x)^2} \, dx = \int |\cos x - \sin x| \, dx

  11. For simplicity, assume cos⁡x−sin⁡x≥0\cos x - \sin x \ge 0 over the interval of integration, allowing us to remove the absolute value.

    ∫(cos⁡x−sin⁡x) dx\int (\cos x - \sin x) \, dx

  12. Integrate each term.

    ∫cos⁡x dx−∫sin⁡x dx\int \cos x \, dx - \int \sin x \, dx

  13. Perform the integration.

    ∫cos⁡x dx=sin⁡x\int \cos x \, dx = \sin x ∫sin⁡x dx=−cos⁡x\int \sin x \, dx = -\cos x

  14. Combine the results and add the constant of integration.

    sin⁡x−(−cos⁡x)+C=sin⁡x+cos⁡x+C\sin x - (-\cos x) + C = \sin x + \cos x + C

    The final answer is sin⁡x+cos⁡x+C\boxed{\sin x + \cos x + C}.

    Q13. Compute the integral: ∫sin⁡2x1+cos⁡x dx\int \frac{\sin^2 x}{1+\cos x} \, dx

  15. Use the trigonometric identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x.

    ∫1−cos⁡2x1+cos⁡x dx\int \frac{1 - \cos^2 x}{1+\cos x} \, dx

  16. Factor the numerator using the difference of squares formula a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

    1−cos⁡2x=(1−cos⁡x)(1+cos⁡x)1 - \cos^2 x = (1 - \cos x)(1 + \cos x)

  17. Substitute the factored numerator back into the integral and simplify.

    ∫(1−cos⁡x)(1+cos⁡x)1+cos⁡x dx=∫(1−cos⁡x) dx\int \frac{(1 - \cos x)(1 + \cos x)}{1+\cos x} \, dx = \int (1 - \cos x) \, dx

  18. Integrate each term.

    ∫1 dx−∫cos⁡x dx\int 1 \, dx - \int \cos x \, dx

  19. Perform the integration.

    ∫1 dx=x\int 1 \, dx = x ∫cos⁡x dx=sin⁡x\int \cos x \, dx = \sin x

  20. Combine the results and add the constant of integration.

    x−sin⁡x+Cx - \sin x + C

    The final answer is x−sin⁡x+C\boxed{x - \sin x + C}.

Was this helpful?

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QSimplify the integral: ∫(3sin x / cos x + π/x + 7) dxQEvaluate the integral of (cos⁻¹ x)2 with respect to x.QEvaluate the integral: integral (dx)/(1-x2)QIntegrate integral (2)/( x) dx
Browse by topic:CalculusTrigonometry
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp