Thank you for uploading the image. It contains four integrals labeled (a), (b), (c), and (d). Let us solve each one step-by-step.

(a) $\int \frac{\sin x \cos x}{\sqrt{\sin x + 1}} \, dx$

Hint: Rewrite the numerator as $(\sin x + 1)\cos x - \cos x$.

1. Rewrite the numerator:
   $\sin x \cos x = \big((\sin x + 1)\cos x - \cos x\big)$ Thus, the integral becomes:
   $\int \frac{(\sin x + 1)\cos x}{\sqrt{\sin x + 1}} \, dx - \int \frac{\cos x}{\sqrt{\sin x + 1}} \, dx$

2. Substitute $u = \sin x + 1$, so $du = \cos x \, dx$:

   • First term: $\int \frac{u}{\sqrt{u}} \, du = \int u^{1/2} \, du = \frac{2}{3} u^{3/2}$.
   • Second term: $\int u^{-1/2} \, du = 2\sqrt{u}$.

3. Back-substitute $u = \sin x + 1$:
   $\frac{2}{3} (\sin x + 1)^{3/2} - 2(\sin x + 1)^{1/2} + C.$

(b) $\int x^{-1/5} \sec\left(x^{4/5}\right) \tan\left(x^{4/5}\right) \, dx$

1. Let $u = x^{4/5}$, so $du = \frac{4}{5}x^{-1/5} \, dx$, or $x^{-1/5} \, dx = \frac{5}{4} du$.

2. The integral becomes:
   $\frac{5}{4} \int \sec u \tan u \, du$

3. Simplify:
   $\frac{5}{4} \sec u + C$

4. Back-substitute $u = x^{4/5}$:
   $\frac{5}{4} \sec\left(x^{4/5}\right) + C.$

(c) $\int \frac{\cos 2x}{(1 + \sin 2x)^2} \, dx$

1. Let $u = 1 + \sin 2x$, so $du = 2 \cos 2x \, dx$, or $\cos 2x \, dx = \frac{1}{2} du$.

2. The integral becomes:
   $\frac{1}{2} \int u^{-2} \, du = \frac{1}{2} \cdot \frac{-1}{u} = -\frac{1}{2u} + C.$

3. Back-substitute $u = 1 + \sin 2x$:
   $-\frac{1}{2(1 + \sin 2x)} + C.$

(d) $\int \frac{1}{(1 + \sqrt{x})^3} \, dx$

1. Let $u = 1 + \sqrt{x}$, so $\sqrt{x} = u - 1$ and $dx = 2(u - 1) \, du$.

2. Substitute into the integral:
   $\int \frac{1}{u^3} \cdot 2(u - 1) \, du = 2 \int \left(\frac{1}{u^2} - \frac{1}{u^3}\right) \, du.$

3. Simplify each term:

   • $\int \frac{1}{u^2} \, du = -\frac{1}{u}$.
   • $\int \frac{1}{u^3} \, du = -\frac{1}{2u^2}$.

4. Combine results:
   $2\left(-\frac{1}{u} + \frac{1}{2u^2}\right) = -\frac{2}{u} + \frac{1}{u^2}.$

5. Back-substitute $u = 1 + \sqrt{x}$:
   $-\frac{2}{1 + \sqrt{x}} + \frac{1}{(1 + \sqrt{x})^2} + C.$

Let me know if you'd like further explanations!

Here are five related questions:

1. Can you solve a similar integral involving a cube root substitution?
2. How do substitutions involving trigonometric identities simplify integrals?
3. What are some key techniques for tackling integrals with square roots?
4. How does integration by parts differ from substitution in application?
5. Why is back-substitution critical after using $u$-substitution?

Tip: Always double-check the substitution to ensure it simplifies all components of the integral!