In Problems \( 27-60 \), find each indefinite integral. \( \begin{array}{ll}\text { (33) 27. } \int \sin (3 x) d x & \text { 28. } \int x \sin x^{2} d x \\ \text { 29. } \int \sin x \cos ^{2} x d x & \text { 30. } \int \tan ^{2} x \sec ^{2} x d x \\ \text { 31. } \int \frac{e^{1 / x}}{x^{2}} d x & \text { 32. } \int \frac{e^{\frac{3}{x}}}{\sqrt[3]{x^{2}}} d x \\ \text { (435) 33. } \int \frac{x d x}{x^{2}-1} & \text { 34. } \int \frac{5 x d x}{1-x^{2}} \\ \text { 35. } \int \frac{e^{x}}{\sqrt{1+e^{x}}} d x & \text { 36. } \int \frac{d x}{x(\ln x)^{7}}\end{array} \)
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To solve the integral \( \int \sin(3x) \, dx \), we can use a basic u-substitution. Let \( u = 3x \), then \( du = 3\,dx \) or \( dx = \frac{du}{3} \). This turns the integral into \( \frac{1}{3} \int \sin(u) \, du \), which evaluates to \( -\frac{1}{3} \cos(u) + C \). Substituting back for \( u \), we find that the result is \( -\frac{1}{3} \cos(3x) + C \). For the integral \( \int x \sin(x^2) \, dx \), again, we can use the u-substitution method. Let \( u = x^2 \) so that \( du = 2x \, dx \) which means \( x \, dx = \frac{du}{2} \). The integral then becomes \( \frac{1}{2} \int \sin(u) \, du \), which evaluates to \( -\frac{1}{2} \cos(u) + C \). Replacing \( u \) with \( x^2 \) again leads us to the final result of \( -\frac{1}{2} \cos(x^2) + C \).