Question

Evaluate:$\int \left({\sin }^{6}x+{\cos }^{6}x+3{\sin }^{2}x{\cos }^{2}x\right)dx$

• A. $x+c$
• B. $\frac{3}{2}\sin 2x+c$
• C. $-\frac{3}{2}\cos 2x+c$
• D. $\frac{1}{3}\sin 3x–\cos 3x+c$
• E. $\frac{1}{3}s\mathrm{in}3x+\cos 3x+c$

Answer 1

The correct option is A

$x+c$

Evaluating the integral

$\int \left({\sin }^{6}x+{\cos }^{6}x+3{\sin }^{2}x{\cos }^{2}x\right)dx$

$$\begin{gathered} \Rightarrow \int \left[{\left({\sin }^{2}x\right)}^3+{\left({\cos }^{2}x\right)}^3+3{\sin }^{2}x{\cos }^{2}x\right]dx \\ \Rightarrow \int \left[\left({\sin }^{2}x+{\cos }^{2}x\right)+\left({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x+3{\sin }^{2}x{\cos }^{2}x\right)\right]dx\left[\because a^3+b^3=\left(a^2+b^2\right)(a^2+b^2-ab)\right] \\ \Rightarrow \int \left({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x\right)dx\left[\because {\sin }^{2}x+{\cos }^{2}x=1\right] \\ \Rightarrow \int {\left({\sin }^{2}x+{\cos }^{2}x\right)}^{2}dx\left[\because {(a+b)}^2=a^2+b^2+2ab\right] \\ \Rightarrow \int dx=x+c\left[cisanintegratingconstsnt\right] \end{gathered}$$

Hence, option A is the correct answer.