Trigonometric Integrals

Approach: $\cos^2 x$ is not one of the basic integration formulas, so rewrite it with the identity $\cos^2 x = \frac{1+\cos 2x}{2}$ as a constant plus a single trigonometric term, and integrate the two parts separately.

Steps:

1. Rewrite with the power-reduction formula: $\cos^2 x = \frac{1 + \cos 2x}{2}$.
2. Substitute and use linearity: $\int \cos^2 x \, dx = \int \frac{1}{2} \, dx + \frac{1}{2}\int \cos 2x \, dx$.
3. Integrate each part: $\int \frac{1}{2} \, dx = \frac{x}{2}$ and $\frac{1}{2}\int \cos 2x \, dx = \frac{1}{2} \cdot \frac{1}{2}\sin 2x = \frac{1}{4}\sin 2x$.
4. Combining gives $\int \cos^2 x \, dx = \frac{x}{2} + \frac{1}{4}\sin 2x + C$.
5. Verify by differentiating: $\frac{d}{dx}\left(\frac{x}{2} + \frac{1}{4}\sin 2x\right) = \frac{1}{2} + \frac{1}{2}\cos 2x = \frac{1 + \cos 2x}{2} = \cos^2 x$, which matches the original integrand.

Answer: $\int \cos^2 x \, dx = \frac{x}{2} + \frac{1}{4}\sin 2x + C$.

Approach: The power-reduction formula cannot be applied directly to an odd power, because reducing it produces terms such as $\cos 2x$ that are not integrable by the basic formulas. The right move is to use the identity $\sin^2 x + \cos^2 x = 1$ to keep one factor of $\sin x$, rewrite the rest in terms of $\cos x$, and then substitute $u = \cos x$.

Steps:

1. Rewrite with $\sin^2 x = 1 - \cos^2 x$: $\sin^3 x = \sin x \cdot \sin^2 x = \sin x (1 - \cos^2 x)$.
2. Let $u = \cos x$, so $du = -\sin x \, dx$.
3. Substitute and integrate: $\int \sin^3 x \, dx = \int \sin x (1 - \cos^2 x) \, dx = -\int (1 - u^2) \, du = -u + \frac{u^3}{3} + C$.
4. Back-substitute $u = \cos x$: $\int \sin^3 x \, dx = -\cos x + \frac{1}{3}\cos^3 x + C$.
5. Verify by differentiating: $\frac{d}{dx}\left(-\cos x + \frac{1}{3}\cos^3 x\right) = \sin x - \cos^2 x \sin x = \sin x (1 - \cos^2 x) = \sin^3 x$, which matches the original integrand.

Answer: $\int \sin^3 x \, dx = -\cos x + \frac{1}{3}\cos^3 x + C$.

Approach: The denominator has the form $a^2+x^2$, so put $x = a\tan t$ and use the identity $1 + \tan^2 t = \sec^2 t$ to turn the denominator into a perfect square, which then cancels against the differential. Taking $t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ gives $\sec t > 0$, so no absolute value is needed for $\sqrt{a^2+x^2} = a\sec t$.

Steps:

1. Let $x = a\tan t$ with $t \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so $dx = a\sec^2 t \, dt$ and $a^2 + x^2 = a^2(1 + \tan^2 t) = a^2 \sec^2 t$.
2. Substitute: $\int \frac{dx}{a^2+x^2} = \int \frac{a\sec^2 t}{a^2 \sec^2 t} \, dt = \frac{1}{a}\int dt = \frac{t}{a} + C$.
3. Back-substitute: from $\tan t = \frac{x}{a}$ we get $t = \arctan\frac{x}{a}$.
4. Verify by differentiating: $\frac{d}{dx}\left[\frac{1}{a}\arctan\frac{x}{a}\right] = \frac{1}{a} \cdot \frac{1/a}{1 + x^2/a^2} = \frac{1}{a^2 + x^2}$, which matches the original integrand.

Answer: $\int \frac{dx}{a^2+x^2} = \frac{1}{a}\arctan\frac{x}{a} + C$.

Approach: The denominator $1+\cos x$ is turned into $2\cos^2\frac{x}{2}$ by the half-angle formula $\cos x = 2\cos^2\frac{x}{2} - 1$, so the integrand becomes $\frac{1}{2}\sec^2\frac{x}{2}$ and can be integrated like a basic formula. (The general method mentioned on this page for $\int \frac{dx}{a+b\cos x}$ is the universal substitution $t = \tan\frac{x}{2}$; for $a = b = 1$ the half-angle formula is more direct.)

Steps:

1. By the half-angle formula, $1 + \cos x = 2\cos^2\frac{x}{2}$, so $\frac{1}{1+\cos x} = \frac{1}{2}\sec^2\frac{x}{2}$.
2. Let $u = \frac{x}{2}$, so $du = \frac{1}{2}dx$ and therefore $dx = 2\,du$.
3. Substitute and integrate: $\int \frac{dx}{1+\cos x} = \frac{1}{2}\int \sec^2 u \cdot 2\,du = \int \sec^2 u \, du = \tan u + C$.
4. Back-substitute $u = \frac{x}{2}$: the original integral is $\tan\frac{x}{2} + C$.
5. Verify by differentiating: $\frac{d}{dx}\tan\frac{x}{2} = \frac{1}{2}\sec^2\frac{x}{2} = \frac{1}{2\cos^2\frac{x}{2}} = \frac{1}{1+\cos x}$, which matches the original integrand.

Answer: $\int \frac{dx}{1+\cos x} = \tan\frac{x}{2} + C$.