Question

Evaluate the following integrals: (i) \(\int \frac{5 \cos ^{3} x+3 \sin ^{3} x}{\sin ^{2} x \cos ^{2} x} d x\) (ii) \(\int\left(\cos ^{6} x+\sin ^{6} x\right) d x\) (iii) \(\int \sin ^{3} x \cos \frac{x}{2} d x\) (iv) \(\int \frac{d x}{\sqrt{3} \cos x+\sin x}\)

Short answer

Question: Evaluate the following integrals: (i) \(\int \frac{5\cos x+3\sin x}{\sin^2 x\cos^2 x} dx\) (ii) \(\int \left(\cos^6 x+\sin^6 x\right)dx\) (iii) \(\int \sin^3x\cos\frac{x}{2}dx\) (iv) \(\int \frac{dx}{\sqrt{3}\cos x+\sin x}\) Solution: (i) \(5\ln|\sec x| - 3\ln|\csc x+\cot x| + C\) (ii) \(\frac{3}{16}\sin 4x - \frac{3}{8}\sin 2x +x +\frac{1}{32}\sin 6x +C\) (iii) \(2\cos^2\frac{x}{2} - \frac{1}{2}\cos^4\frac{x}{2} + C\) (iv) \(\frac{1}{2}\tan^2x + \ln|\sqrt{3}\tan x| + C\)

Step-by-step solution

Simplify the integrand

Divide both the numerator and denominator of the integrand function by \(\sin^2x\cos^2x\). We get $$\int \frac{\frac{5\cos x}{\sin x \cos x} + \frac{3\sin x}{\sin x \cos x}}{1}dx.$$ Now, simplifying gives $$\int (5\tan x + 3\csc x) dx.$$

Evaluate the integral

Now we have $$\int (5\tan x + 3\csc x) dx = 5\int \tan x dx + 3\int \csc x dx.$$ Using known antiderivatives, we get the final result $$5\ln|\sec x| - 3\ln |\csc x+\cot x| + C.$$ (ii)

Use a trigonometric identity

Rewrite the integral using the identity \(\sin^2x+\cos^2x=1\) and the fact \((a+b)^3=a^3+b^3+3ab(a+b)\), we get $$\int\left(\cos ^{6} x+\sin ^{6} x\right) d x = \int \left(\cos^2x\right)^3dx + \int \left(\sin^2x\right)^3 dx + \int 3\sin^2x\cos^2x\left(\sin^2x+\cos^2x\right)dx$$

Evaluate the integral

Integrating the above expression, we get $$\frac{3}{16}\sin 4x - \frac{3}{8}\sin 2x +x +\frac{1}{32}\sin 6x +C.$$ (iii)

Apply substitution

Let \(u = \cos\frac{x}{2}\). Then \(du = \frac{-1}{2}\sin\frac{x}{2}dx\). In addition, notice that \(\sin^3x=(\sin^2x)(\sin x)=(1-\cos^2x)(2\sin\frac{x}{2}\cos\frac{x}{2})\).

Evaluate the integral

Now we write the integral in terms of \(u\): $$\int \sin ^{3} x \cos \frac{x}{2} d x = 2\int (1-u^2)(2u)du.$$ $$= 4\int(u-u^3)du=4\left(\frac{1}{2}u^2-\frac{1}{4}u^4\right)+C$$ Substituting back \(u = \cos\frac{x}{2}\) gives the final result $$2\cos^2\frac{x}{2} - \frac{1}{2}\cos^4\frac{x}{2} + C.$$ (iv)

Apply substitution

Let \(u=\sqrt{3}\tan x\). Then \(du=\sqrt{3}\sec^2x dx\). Note that \(\frac{1}{\cos^2x}=\sec^2x=1+\tan^2x=1+\frac{u^2}{3}\).

Evaluate the integral

Now we write the integral in terms of \(u\): $$\int \frac{dx}{\sqrt{3}\cos x+\sin x} = \frac{1}{\sqrt{3}}\int \frac{\sec^2x dx}{u+3} = \frac{1}{\sqrt{3}}\int \frac{1+\frac{u^2}{3}}{u}du.$$ Rewriting the above expression $$\int \frac{u^2+3}{3u} du = \frac{1}{\sqrt{3}}\int\left(\frac{u^2}{3u} + \frac{3}{u}\right)du$$ Now, integrate the expression: $$\frac{1}{\sqrt{3}}\left(\frac{1}{3}\int udu + 3\int \frac{1}{u}du\right) = \frac{1}{\sqrt{3}}\left(\frac{1}{6}u^2 + 3\ln |u|\right) + C.$$ Substituting back \(u=\sqrt{3}\tan x\) gives the final result: $$\frac{1}{2}\tan^2x + \ln|\sqrt{3}\tan x| + C.$$