Question

Which of the following statements are true with respect to definite integrals ?

• A. ${\int }_0^{\frac{\pi }{2}}co{s}^{m}xsi{n}^{m}xdx=2{\int }_0^{\frac{\pi }{2}}(cosx{)}^{m}dx,mϵN$
• B. ${\int }_0^{\frac{\pi }{2}}\frac{xsinx}{1+co{s}^{2}x}dx=\frac{\pi }{2}{\int }_0^1\frac{dx}{1+x^2}$
• C. ${\int }_0^{\frac{\pi }{2}}\frac{cos3x+1}{2cosx-1}dx={\int }_0^{\frac{\pi }{2}}\frac{1-cos3x}{1+2cosx}dx$
• D. ${\int }_0^{\frac{\pi }{2}}\left(\frac{1+sin3x}{1+2sinx}\right)dx={\int }_0^{\frac{\pi }{2}}\frac{cos3x+1}{2cosx-1}dx$

Answer 1

Answer: (A), (C), (D)

The correct options are
A ${\int }_0^{\frac{\pi }{2}}co{s}^{m}xsi{n}^{m}xdx=2{\int }_0^{\frac{\pi }{2}}(cosx{)}^{m}dx,mϵN$
C ${\int }_0^{\frac{\pi }{2}}\frac{cos3x+1}{2cosx-1}dx={\int }_0^{\frac{\pi }{2}}\frac{1-cos3x}{1+2cosx}dx$
D ${\int }_0^{\frac{\pi }{2}}\left(\frac{1+sin3x}{1+2sinx}\right)dx={\int }_0^{\frac{\pi }{2}}\frac{cos3x+1}{2cosx-1}dx$

(a) $L.H.S=2^{-m}{\int }_0^{\frac{\pi }{2}}(sin2x{)}^{m}dx=\frac{2^{-m}}{2}{\int }_0^{\pi }(sint{)}^{m}dt$
$=2^{-m}{\int }_0^{\frac{\pi }{2}}(sint{)}^{m}dt=2^{-m}{\int }_0^{\frac{\pi }{2}}(cost{)}^{m}dt=RHS$
(b) $I={\int }_0^{\pi }\frac{(\pi -x)sinx}{1+co{s}^{2}x}dx\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx$
$\Rightarrow 2I=2\pi {\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+co{s}^{2}x}dx$
$\Rightarrow I=\pi {\int }_0^{\frac{\pi }{2}}\frac{sinx}{1+co{s}^{2}x}dx=\pi {\int }_0^1\frac{dt}{1+t^2}\ne \frac{\pi }{2}{\int }_0^1\frac{dt}{1+t^2}$
(c) ${\int }_0^{\frac{\pi }{2}}\frac{1-cos3x}{1+2cosx}dx={\int }_0^{\frac{\pi }{2}}\frac{1+cos3x}{2cosx-1}dx=1$
(d) $I={\int }_0^{\frac{\pi }{2}}\frac{1+sin3x}{1+2sinx}dx$
$Now,using{\int }_0^{a}f(a-x)dx={\int }_0^{a}f\left(x\right)dx$
${\int }_0^{\frac{\pi }{2}}\frac{1+cos3x}{2cosx-1}dx={\int }_0^{\frac{\pi }{2}}(cos2x+cosx)dx=1$
$\therefore {\int }_0^{\frac{\pi }{2}}\frac{1+sin3x}{1+2sinx}dx={\int }_0^{\frac{\pi }{2}}\frac{1+cos3x}{2cosx-1}dx$