If 1a^2 ^2x+b^2 ^2xdx=(1)/(12) ^-1(3 x)+ constant, then the maximum value of a x+b x, is :

Question

If $\int \frac{1}{{a}^{2}{\sin }^{2}x+{b}^{2}{\cos }^{2}x}dx=\frac{1}{12}{	an }^{-1}(3	an x)+$ constant, then the maximum value of $a\sin x+b\cos x$, is :

Accepted Answer

$\begin{gathered} \int \frac{1}{{a}^{2}{\sin }^{2}x+{b}^{2}{\cos }^{2}x}dx=\frac{1}{12}{	an }^{-1}(3	an x)+c \ I=\int \frac{{\sec }^{2}x}{{b}^{2}+{a}^{2}{	an }^{2}x}dx \ 	an x=t \ \Rightarrow {\sec }^{2}xdx=dt \ I=\int \frac{dt}{{b}^{2}+{a}^{2}{t}^{2}} \ =\frac{1}{ba}{	an }^{-1}(\frac{at}{b})+c \ I=\frac{1}{ab}{	an }^{-1}(\frac{a}{b}	an x)+c \ \Rightarrow ab=12\frac{a}{b}=3 \ \Rightarrow {a}^{2}=36{b}^{2}=4 \ \Rightarrow a\sin x+b\cos x\sqrt{{a}^{2}+{b}^{2}} \end{gathered}$

$\, =\sqrt{36+4}$

$\, =\sqrt{40}$

Solutions · 1