Question

Integrate x/ a²cos²x+ b²sin²x

Answer 1

Answer:

I=∫0πxa2cos2x+b2sin2xdx  ......... (1)

=>I=∫0ππ−xa2cos2(π−x)+b2sin2(π−x)dx     [Using Property,∫0af(x)dx=∫0af(a−x)dx]

=>I=∫0ππa2cos2x+b2sin2xdx−∫0πxa2cos2x+b2sin2xdx ....... (2)

Adding (1) and (2), We Get  

=>2I=∫0ππa2cos2x+b2sin2xdx        

=>I=π2∫0π1a2cos2x+b2sin2xdx

=>I=2.π2∫0π21a2cos2x+b2sin2xdx          [Using Property,∫02af(x)dx=2∫0af(2a−x)dx,if f(2a−x)=f(x)]

=>I=π∫0π21a2cos2x+b2sin2xdx

Divinde Numerator and Denominator by cos2x, We get  

=>I=π∫0π21cos2xa2cos2xcos2x+b2sin2xcos2xdx      

=>I=π∫0π2sec2xa2+b2tan2xdx

Put b.tanx=t,then b.sec2xdx=dt, sec2x.dx=1bdt  

also, when x=0, t=0 and when x=π2,t→∞

So,  

=>I=πb∫0∞1a2+t2dt          

=>I=πb.1a|tan−1(ta)|∞0

=>I=πab[π2−0] = π22ab

Step-by-step explanation:

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