MCQ
$\int_0^\pi {\frac{{x\,\tan x}}{{\sec x + \cos x}}} \,dx = $
- ✓$\frac{{{\pi ^2}}}{4}$
- B$\frac{{{\pi ^2}}}{2}$
- C$\frac{{3{\pi ^2}}}{2}$
- D$\frac{{{\pi ^2}}}{3}$
It gives $I = \frac{\pi }{2}\int_0^\pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} dx$
Now put $\cos x = t$ and solve, we get
$I = \frac{\pi }{2} \times \frac{\pi }{2} = \frac{{{\pi ^2}}}{4}$.
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$(S_1)$ $\mathrm{x}_{1}, \mathrm{x}_{2} \in(2,4), \mathrm{x}_{1}<\mathrm{x}_{2}$ અસ્તિત્વ ધરાવે કે જેથી $f^{\prime}\left(x_{1}\right)=-1$ અને $f^{\prime}\left(x_{2}\right)=0$
$(S_2)$ $\mathrm{x}_{3}, \mathrm{x}_{4} \in(2,4), \mathrm{x}_{3}<\mathrm{x}_{4}$, અસ્તિત્વ ધરાવે કે જેથી $f$ એ $\left(2, x_{4}\right)$ માં ઘટતું વિધેય, $\left(x_{4}, 4\right)$ માં વધતું વિધેય અને $2 f^{\prime}\left(x_{3}\right)=\sqrt{3} f\left(x_{4}\right)$ થાય. તો . .. .