I = ∫ t^2 dt; I = ∫ t + 2 dt; I = -1/t + C; I = -1/(cos x - sin x); I = ∫ sin^3 x + cos^2 x dx; I = ∫ tan x sec x + tan x cot x cos x dx; I = ∫ sin(cos x) dx, and more.

Steps to Solve

1. Integrate each expression separately

We can integrate each expression in the integral one by one.

2. First integral: $I_1 = \int t^2 , dt$

Using the power rule of integration: $$ I_1 = \frac{t^{3}}{3} + C_1 $$

3. Second integral: $I_2 = \int (t + 2) , dt$

Split the integral: $$ I_2 = \int t , dt + \int 2 , dt $$ Integrating each part: $$ I_2 = \frac{t^{2}}{2} + 2t + C_2 $$

4. Third integral: $I_3 = \int \frac{-1}{t} , dt$

This integral results in: $$ I_3 = -\ln |t| + C_3 $$

5. Fourth integral: $I_4 = \int \frac{-1}{\cos x - \sin x} , dx$

This is a bit complex; we can use substitution or a direct integration approach. Letting $u = \cos x - \sin x$, gives: $$ I_4 = -\ln |\cos x - \sin x| + C_4 $$

6. Fifth integral: $I_5 = \int (\sin^3 x + \cos^2 x) , dx$

We can break it down:

• For $\int \sin^3 x , dx$ we can use the identity $\sin^3 x = (1 - \cos^2 x)\sin x$ and apply substitution.
• For $\int \cos^2 x , dx$, we can apply the identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$.

After integration: $$ I_5 = -\frac{1}{3}\cos^3 x + \frac{x}{2} + \frac{1}{4}\sin(2x) + C_5 $$

7. Sixth integral: $I_6 = \int (\tan x \sec x + \tan x \cot x \cos x) , dx$

This can be simplified: $$ I_6 = \int \tan x \sec x , dx + \int \tan x , dx $$ Using the fact that $\int \tan x \sec x , dx = \sec x$ and $\int \tan x , dx = -\ln |\cos x|$, we get: $$ I_6 = \sec x - \ln |\cos x| + C_6 $$

8. Seventh integral: $I_7 = \int \sin(\cos x) , dx$

This integral doesn’t have a simple antiderivative and generally would be evaluated numerically or with series expansions unless specified otherwise: $$ I_7 \text{ does not have a closed form.} $$