MCQ
If $f(x)\, = sin\, (sin\,x)$ and $f"(x) + tan\,xf'(x) + g(x)\, = 0$, then $g(x)$ is
- A$cos^2\, x\, cos\, (sin\, x)$
- B$sin^2\,x\, cos\, (cos\,x)$
- C$sin^2\, x\, sin (cos\,x)$
- ✓$cos^2\, x\, sin\, (sin\,x)$
$ \Rightarrow f'\left( x \right) = \cos \left( {\sin x} \right).\cos x$
$ \Rightarrow f''\left( x \right) = - \sin \left( {\sin x} \right).{\cos ^2}x + \cos \left( {\sin x} \right).\left( { - \sin x} \right)$
$ = - {\cos ^2}x.\sin \left( {\sin x} \right) - \sin x.\cos \left( {\sin x} \right)$
Now $f''\left( x \right) + \tan x.f'\left( x \right) + g\left( x \right) = 0$
$ \Rightarrow g\left( x \right) = {\cos ^2}x.\sin \left( {\sin x} \right) + \sin x.\cos \left( {\sin x} \right)$
$ - \tan x.\cos x.\cos \left( {\sin x} \right)$
$ \Rightarrow g\left( x \right) = {\cos ^2}x.\sin \left( {\sin x} \right)$
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${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then