Question

Evaluate the following integrals: (i) \(\int \cos x^{\circ} d x\) (ii) \(\int \sec ^{2}(a x+b) d x\) (iii) \(\int \cot ^{2} x d x\) (iv) \(\int \frac{1-\cos x}{1+\cos x} d x\)

Short answer

Question: Find the integrals of the following functions: (i) \(\int \cos x^{\circ} d x\) (ii) \(\int \sec^2(ax+b) dx\) (iii) \(\int \cot^2 x dx\) (iv) \(\int \frac{1-\cos x}{1+\cos x} dx\) Answer: (i) \(\frac{180}{\pi}(\sin \frac{\pi}{180}x + C)\) (ii) \(\frac{1}{a}(\tan(ax+b) + C)\) (iii) \(-\cot x + x + C\) (iv) \(x + C\)

Step-by-step solution

(i) Integral of \(\int \cos x^{\circ} d x\)

In this case, the angle is given in degrees. To make the integration process easier, we'll first convert the angle to radians by multiplying by \(\frac{\pi}{180}\): $$ x^{\circ} =\frac{\pi}{180}x $$ Now, we can find the integral: $$ \int \cos \left(\frac{\pi}{180}x\right) dx $$ Apply the substitution \(u = \frac{\pi}{180}x\), so \(du = \frac{\pi}{180}dx\). This results in: $$ \int \cos u \frac{180}{\pi} du = \frac{180}{\pi} \int \cos u du $$ Now, integrate with respect to \(u\): $$ \frac{180}{\pi} \int \cos u du = \frac{180}{\pi}(\sin u + C) $$ Finally, substitute back the angle in radians: $$ \frac{180}{\pi}(\sin \frac{\pi}{180}x + C) $$

(ii) Integral of \(\int \sec^2(ax+b) dx\)

To solve this integral, integrate by substitution. Make the substitution \(u=ax+b\), so \(du = adx\). $$ \int \sec^2(u) \frac{1}{a} du = \frac{1}{a} \int \sec^2 u du $$ Now, integrate with respect to \(u\): $$ \frac{1}{a} \int \sec^2 u du = \frac{1}{a}(\tan u + C) $$ Substitute back the original expression for \(u\): $$ \frac{1}{a}(\tan(ax+b) + C) $$

(iii) Integral of \(\int \cot^2 x dx\)

To find the integral of \(\cot^2 x\), first recall the identity \(\cot^2 x + 1 = \csc^2 x\). Rewriting the original integral as \(\int (\csc^2 x - 1) dx\), we can integrate each term separately: $$ \int (\csc^2 x - 1)dx = \int \csc^2 x dx - \int 1 dx $$ Integration of each term: $$ \int \csc^2 x dx = -\cot x + C_1 $$ $$ \int 1 dx = x + C_2 $$ Combine the results: $$ -\cot x + x + C_1 + C_2 = -\cot x + x + C $$

(iv) Integral of \(\int \frac{1-\cos x}{1+\cos x} dx\)

Use the trigonometric identity \(\sin^2 x = 1 - \cos^2 x\). Then, rewrite the integral as follows: $$ \int \frac{1-\cos x}{1+\cos x} dx = \int \frac{1-\cos x}{1+\cos x} \frac{1-\cos x}{1-\cos x} dx $$ Simplifying: $$ \int \frac{(1-\cos x)^2}{1-\cos^2 x} dx = \int \frac{\sin^2 x}{1-\cos^2 x} dx = \int \frac{\sin^2 x}{\sin^2 x} dx = \int 1 dx $$ Now, integrate with respect to \(x\): $$ \int 1 dx = x + C $$