1.) ∫ (sin x) / (cos^2 x) dx

2.) ∫ (1) / (1+x^2) dx

3.) ∫ xe^x^2 dx

4.) ∫ x^2 sinx dx

5.) ∫ (lnx) / (x) dx

Most of these can be solved by the method of variable substitution. Define one function of x to be "u" and then calculate du.

For example, in #3, let u = x^2 so that
du = 2x dx. Then

�ç xe^x^2 dx = (1/2)�ç e^u du

= (1/2) e^u = (1/2) e^(x^2)

See how many of the others you can do with that method.

1.) To evaluate the integral ∫(sin x) / (cos^2 x) dx, we can use a substitution. Let u = cos x, so du = -sin x dx. Rewriting the integral in terms of u gives us ∫-(1 / u^2) du. This is a simple power rule integral, so we can evaluate it as -(1 / -1)u^(-1) + C = u^(-1) + C = 1 / cos x + C.

2.) To evaluate the integral ∫(1) / (1+x^2) dx, we recognize that the integrand resembles the derivative of arctan x. Using the substitution u = x, we get du = dx. Rewriting the integral in terms of u gives us ∫(1) / (1+u^2) du. This is the integral of arctan u, which is tan^(-1) u + C. Making the substitution back to x, we have the final result as tan^(-1) x + C.

3.) To evaluate the integral ∫xe^(x^2) dx, we can use a substitution. Let u = x^2, so du = 2x dx. Rewriting the integral in terms of u gives us (1/2) * ∫e^u du. This is a straightforward integral, and the result is (1/2) * e^u + C = (1/2) * e^(x^2) + C.

4.) To evaluate the integral ∫x^2sinx dx, we can use integration by parts. Let u = x^2 and dv = sinx dx. Differentiating u gives du = 2x dx, and integrating dv gives v = -cosx. Applying the integration by parts formula, ∫u dv = uv - ∫v du, gives us x^2(-cosx) - ∫(-cosx)(2x dx). Simplifying, we have -x^2cosx + 2∫xcosx dx. This integral can be evaluated again using integration by parts with u = x and dv = cosx dx. Differentiating u gives du = dx, and integrating dv gives v = sinx. Applying the formula again, we get -x^2cosx + 2(xsinx - ∫sinx dx). The integral of sinx is -cosx, so our final result is -x^2cosx + 2(xsinx + cosx) + C.

5.) To evaluate the integral ∫(lnx) / x dx, we can use integration by parts. Let u = lnx, so du = (1/x) dx. Differentiating u gives dv = dx. Integrating dv gives v = x. Applying the integration by parts formula, ∫u dv = uv - ∫v du, gives us xlnx - ∫x * (1/x) dx. Simplifying, we have xlnx - ∫dx. The integral of dx is x, so our final result is xlnx - x + C.