JEE Main 2014 — Mathematics Calculus

$$\int _{0}^{\pi }\sqrt{1+4 {\text{sin}}^{2}\frac{x}{2}-4 \text{sin}\frac{x}{2}}dx$$∫0π​1+4sin22x​−4sin2x​​dx

$$=\int _{0}^{\pi }\sqrt{{(2 \text{sin}\frac{x}{2}-1)}^{2}}dx$$=∫0π​(2sin2x​−1)2​dx

$$=\int _{0}^{\pi }|2 \text{sin}\frac{x}{2}-1|dx$$=∫0π​∣2sin2x​−1∣dx

$$=2\int _{0}^{\pi }|\text{sin}\frac{x}{2}-\frac{1}{2}|dx$$=2∫0π​∣sin2x​−21​∣dx

$$=2 [\int _{0}^{\pi /3}|\text{sin}\frac{x}{2}-\frac{1}{2}|dx+\int _{\pi /3}^{\pi }|\text{sin}\frac{x}{2}-\frac{1}{2}|dx]$$=2[∫0π/3​∣sin2x​−21​∣dx+∫π/3π​∣sin2x​−21​∣dx]

$$=2 [\int _{0}^{\pi /3}-(\text{sin}\frac{x}{2}-\frac{1}{2})dx+\int _{\pi /3}^{\pi }(\text{sin}\frac{x}{2}-\frac{1}{2})dx]$$=2[∫0π/3​−(sin2x​−21​)dx+∫π/3π​(sin2x​−21​)dx]

$$=2 [{[2\text{cos}\frac{x}{2}+\frac{x}{2}]}_{0}^{\pi /3}+{[-2\text{cos}\frac{x}{2}-\frac{x}{2}]}_{\pi /3}^{\pi }]$$=2[[2cos2x​+2x​]0π/3​+[−2cos2x​−2x​]π/3π​]

$$= 2 [ [ 2 ( \frac{ \sqrt{ 3 } }{ 2 } ) + \frac{ \pi }{ 6 } - 2 ] + [ - \frac{ \pi }{ 2 } + 2 ( \frac{ \sqrt{ 3 } }{ 2 } ) + \frac{ \pi }{ 6 } ] ]$$=2[[2(23​​)+6π​−2]+[−2π​+2(23​​)+6π​]]

$$= 2 [ \sqrt{ 3 } + \frac{ \pi }{ 6 } - 2 - \frac{ \pi }{ 2 } + \sqrt{ 3 } + \frac{ \pi }{ 6 } ]$$=2[3​+6π​−2−2π​+3​+6π​]

$$= 2 [ 2 \sqrt{ 3 } - \frac{ \pi }{ 6 } - 2 ]$$=2[23​−6π​−2]

$= 4 \sqrt{ 3 } - \frac{ \pi }{ 3 } - 4$=43​−3π​−4.