NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86387
Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function and \(\mathrm{f}(1)\) =4. What is the value of \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}\) equal to?
1 \(8 \mathrm{f}^{\prime}(1)\)
2 \(4 \mathrm{f}^{\prime}(1)\)
3 \(2 \mathrm{f}^{\prime}(1)\)
4 \(\mathrm{f}^{\prime}(1)\)
Explanation:
(A) :We have, \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}=\lim _{x \rightarrow 1} \frac{2}{x-1}\left[\frac{t^{2}}{2}\right]_{4}^{f(x)}\) \(=\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}\) Since, \(f(1)=4\) so that limit is \(\left(\frac{0}{0}\right)\) form. On applying L'-Hospital Rule, \(\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}=\lim _{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}\) \(=2 f(1) f^{\prime}(1)=2 \times 4 f^{\prime}(1)=8 f^{\prime}(1)\)
SCRA-2012
Integral Calculus
86388
The integral \(16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}\) is equal to
86389
Let the solution curve \(y=f(x)\) of the differential equation \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)\) pass through the origin. Then \(\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \mathrm{d} x\) is
1 \(\frac{\pi}{3}-\frac{1}{4}\)
2 \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
3 \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)
4 \(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\)
Explanation:
(B) : Given, differential equation is- \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) \(\frac{d y}{d x}-\left(\frac{x}{1-x^{2}}\right) y=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) It is a linear differential equation, Therefore, I.F. \(=\mathrm{e}^{\int\left(\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}\right)}=\mathrm{e}^{\frac{1}{2} \log \left(1-\mathrm{x}^{2}\right)}\) \(=\sqrt{1-\mathrm{x}^{2}}\) Now, multiplying I.F. with equation (i), we get- \(\sqrt{1-x^{2}} \frac{d y}{d x}-\sqrt{1-x^{2}} \frac{x}{\left(1-x^{2}\right)} y=x^{4}+2 x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1-\mathrm{x}^{2}} \mathrm{y}\right)=\mathrm{x}^{4}+2 \mathrm{x}\) \(\int d\left(\sqrt{1-x^{2}} y\right)=\int\left(x^{4}+2 x\right) d x\) \(y \sqrt{1-x^{2}}=\frac{x^{5}}{5}+\frac{2 x^{2}}{2}+C\) Using initial conditions, that is \(f(0)=0\) \(\sqrt{1-\mathrm{x}^{2}}(0)=0+0+\mathrm{C}\) Therefore, \(\mathrm{C}=0\) Therefore, equation (ii) gives \(\sqrt{1-x^{2}} y=\frac{x^{5}}{5}+x^{2}\) Hence, \(y=\frac{\frac{x^{5}}{5}+x^{2}}{\sqrt{1-x^{2}}}=\frac{x^{5}}{5 \sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}\) Now, \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{5}}{5 \sqrt{1-x^{2}}} d x+\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) \(=0+2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) Since even function now solving \(=\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{x}^{2}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Putting \(x=\sin \theta\). Therefore, \(d x=\cos \theta d \theta\) \(\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\sin ^{2} \theta \cos \theta \mathrm{d} \theta}{\cos \theta}=\int_{0}^{\frac{\sqrt{3}}{2}}\left(\frac{1-\cos 2 \theta}{2}\right) \mathrm{d} \theta\) \(=\left[\frac{1}{2} \theta\right]_{0}^{\frac{\pi}{3}}-\frac{1}{2}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{3}}=\frac{1}{2} \times \frac{\pi}{3}-\frac{1}{4}\left[\sin \frac{2 \pi}{3}\right]\) \(=\frac{\pi}{6}-\frac{1}{4} \frac{\sqrt{3}}{2}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) Hence, from Equation (iii), we get - \(2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)=\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
JEE Main-2022-26.07.2022
Integral Calculus
86390
The integral \(\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x\) is equal to :
86387
Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function and \(\mathrm{f}(1)\) =4. What is the value of \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}\) equal to?
1 \(8 \mathrm{f}^{\prime}(1)\)
2 \(4 \mathrm{f}^{\prime}(1)\)
3 \(2 \mathrm{f}^{\prime}(1)\)
4 \(\mathrm{f}^{\prime}(1)\)
Explanation:
(A) :We have, \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}=\lim _{x \rightarrow 1} \frac{2}{x-1}\left[\frac{t^{2}}{2}\right]_{4}^{f(x)}\) \(=\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}\) Since, \(f(1)=4\) so that limit is \(\left(\frac{0}{0}\right)\) form. On applying L'-Hospital Rule, \(\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}=\lim _{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}\) \(=2 f(1) f^{\prime}(1)=2 \times 4 f^{\prime}(1)=8 f^{\prime}(1)\)
SCRA-2012
Integral Calculus
86388
The integral \(16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}\) is equal to
86389
Let the solution curve \(y=f(x)\) of the differential equation \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)\) pass through the origin. Then \(\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \mathrm{d} x\) is
1 \(\frac{\pi}{3}-\frac{1}{4}\)
2 \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
3 \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)
4 \(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\)
Explanation:
(B) : Given, differential equation is- \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) \(\frac{d y}{d x}-\left(\frac{x}{1-x^{2}}\right) y=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) It is a linear differential equation, Therefore, I.F. \(=\mathrm{e}^{\int\left(\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}\right)}=\mathrm{e}^{\frac{1}{2} \log \left(1-\mathrm{x}^{2}\right)}\) \(=\sqrt{1-\mathrm{x}^{2}}\) Now, multiplying I.F. with equation (i), we get- \(\sqrt{1-x^{2}} \frac{d y}{d x}-\sqrt{1-x^{2}} \frac{x}{\left(1-x^{2}\right)} y=x^{4}+2 x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1-\mathrm{x}^{2}} \mathrm{y}\right)=\mathrm{x}^{4}+2 \mathrm{x}\) \(\int d\left(\sqrt{1-x^{2}} y\right)=\int\left(x^{4}+2 x\right) d x\) \(y \sqrt{1-x^{2}}=\frac{x^{5}}{5}+\frac{2 x^{2}}{2}+C\) Using initial conditions, that is \(f(0)=0\) \(\sqrt{1-\mathrm{x}^{2}}(0)=0+0+\mathrm{C}\) Therefore, \(\mathrm{C}=0\) Therefore, equation (ii) gives \(\sqrt{1-x^{2}} y=\frac{x^{5}}{5}+x^{2}\) Hence, \(y=\frac{\frac{x^{5}}{5}+x^{2}}{\sqrt{1-x^{2}}}=\frac{x^{5}}{5 \sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}\) Now, \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{5}}{5 \sqrt{1-x^{2}}} d x+\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) \(=0+2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) Since even function now solving \(=\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{x}^{2}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Putting \(x=\sin \theta\). Therefore, \(d x=\cos \theta d \theta\) \(\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\sin ^{2} \theta \cos \theta \mathrm{d} \theta}{\cos \theta}=\int_{0}^{\frac{\sqrt{3}}{2}}\left(\frac{1-\cos 2 \theta}{2}\right) \mathrm{d} \theta\) \(=\left[\frac{1}{2} \theta\right]_{0}^{\frac{\pi}{3}}-\frac{1}{2}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{3}}=\frac{1}{2} \times \frac{\pi}{3}-\frac{1}{4}\left[\sin \frac{2 \pi}{3}\right]\) \(=\frac{\pi}{6}-\frac{1}{4} \frac{\sqrt{3}}{2}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) Hence, from Equation (iii), we get - \(2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)=\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
JEE Main-2022-26.07.2022
Integral Calculus
86390
The integral \(\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x\) is equal to :
86387
Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function and \(\mathrm{f}(1)\) =4. What is the value of \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}\) equal to?
1 \(8 \mathrm{f}^{\prime}(1)\)
2 \(4 \mathrm{f}^{\prime}(1)\)
3 \(2 \mathrm{f}^{\prime}(1)\)
4 \(\mathrm{f}^{\prime}(1)\)
Explanation:
(A) :We have, \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}=\lim _{x \rightarrow 1} \frac{2}{x-1}\left[\frac{t^{2}}{2}\right]_{4}^{f(x)}\) \(=\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}\) Since, \(f(1)=4\) so that limit is \(\left(\frac{0}{0}\right)\) form. On applying L'-Hospital Rule, \(\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}=\lim _{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}\) \(=2 f(1) f^{\prime}(1)=2 \times 4 f^{\prime}(1)=8 f^{\prime}(1)\)
SCRA-2012
Integral Calculus
86388
The integral \(16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}\) is equal to
86389
Let the solution curve \(y=f(x)\) of the differential equation \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)\) pass through the origin. Then \(\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \mathrm{d} x\) is
1 \(\frac{\pi}{3}-\frac{1}{4}\)
2 \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
3 \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)
4 \(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\)
Explanation:
(B) : Given, differential equation is- \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) \(\frac{d y}{d x}-\left(\frac{x}{1-x^{2}}\right) y=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) It is a linear differential equation, Therefore, I.F. \(=\mathrm{e}^{\int\left(\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}\right)}=\mathrm{e}^{\frac{1}{2} \log \left(1-\mathrm{x}^{2}\right)}\) \(=\sqrt{1-\mathrm{x}^{2}}\) Now, multiplying I.F. with equation (i), we get- \(\sqrt{1-x^{2}} \frac{d y}{d x}-\sqrt{1-x^{2}} \frac{x}{\left(1-x^{2}\right)} y=x^{4}+2 x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1-\mathrm{x}^{2}} \mathrm{y}\right)=\mathrm{x}^{4}+2 \mathrm{x}\) \(\int d\left(\sqrt{1-x^{2}} y\right)=\int\left(x^{4}+2 x\right) d x\) \(y \sqrt{1-x^{2}}=\frac{x^{5}}{5}+\frac{2 x^{2}}{2}+C\) Using initial conditions, that is \(f(0)=0\) \(\sqrt{1-\mathrm{x}^{2}}(0)=0+0+\mathrm{C}\) Therefore, \(\mathrm{C}=0\) Therefore, equation (ii) gives \(\sqrt{1-x^{2}} y=\frac{x^{5}}{5}+x^{2}\) Hence, \(y=\frac{\frac{x^{5}}{5}+x^{2}}{\sqrt{1-x^{2}}}=\frac{x^{5}}{5 \sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}\) Now, \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{5}}{5 \sqrt{1-x^{2}}} d x+\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) \(=0+2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) Since even function now solving \(=\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{x}^{2}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Putting \(x=\sin \theta\). Therefore, \(d x=\cos \theta d \theta\) \(\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\sin ^{2} \theta \cos \theta \mathrm{d} \theta}{\cos \theta}=\int_{0}^{\frac{\sqrt{3}}{2}}\left(\frac{1-\cos 2 \theta}{2}\right) \mathrm{d} \theta\) \(=\left[\frac{1}{2} \theta\right]_{0}^{\frac{\pi}{3}}-\frac{1}{2}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{3}}=\frac{1}{2} \times \frac{\pi}{3}-\frac{1}{4}\left[\sin \frac{2 \pi}{3}\right]\) \(=\frac{\pi}{6}-\frac{1}{4} \frac{\sqrt{3}}{2}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) Hence, from Equation (iii), we get - \(2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)=\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
JEE Main-2022-26.07.2022
Integral Calculus
86390
The integral \(\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x\) is equal to :
86387
Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be a differentiable function and \(\mathrm{f}(1)\) =4. What is the value of \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}\) equal to?
1 \(8 \mathrm{f}^{\prime}(1)\)
2 \(4 \mathrm{f}^{\prime}(1)\)
3 \(2 \mathrm{f}^{\prime}(1)\)
4 \(\mathrm{f}^{\prime}(1)\)
Explanation:
(A) :We have, \(\lim _{x \rightarrow 1} \int_{4}^{f(x)} \frac{2 t d t}{x-1}=\lim _{x \rightarrow 1} \frac{2}{x-1}\left[\frac{t^{2}}{2}\right]_{4}^{f(x)}\) \(=\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}\) Since, \(f(1)=4\) so that limit is \(\left(\frac{0}{0}\right)\) form. On applying L'-Hospital Rule, \(\lim _{x \rightarrow 1} \frac{[f(x)]^{2}-16}{x-1}=\lim _{x \rightarrow 1} \frac{2 f(x) f^{\prime}(x)}{1}\) \(=2 f(1) f^{\prime}(1)=2 \times 4 f^{\prime}(1)=8 f^{\prime}(1)\)
SCRA-2012
Integral Calculus
86388
The integral \(16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}\) is equal to
86389
Let the solution curve \(y=f(x)\) of the differential equation \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}, x \in(-1,1)\) pass through the origin. Then \(\int_{\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) \mathrm{d} x\) is
1 \(\frac{\pi}{3}-\frac{1}{4}\)
2 \(\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
3 \(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)
4 \(\frac{\pi}{6}-\frac{\sqrt{3}}{2}\)
Explanation:
(B) : Given, differential equation is- \(\frac{d y}{d x}+\frac{x y}{x^{2}-1}=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) \(\frac{d y}{d x}-\left(\frac{x}{1-x^{2}}\right) y=\frac{x^{4}+2 x}{\sqrt{1-x^{2}}}\) It is a linear differential equation, Therefore, I.F. \(=\mathrm{e}^{\int\left(\frac{-\mathrm{x}}{1-\mathrm{x}^{2}}\right)}=\mathrm{e}^{\frac{1}{2} \log \left(1-\mathrm{x}^{2}\right)}\) \(=\sqrt{1-\mathrm{x}^{2}}\) Now, multiplying I.F. with equation (i), we get- \(\sqrt{1-x^{2}} \frac{d y}{d x}-\sqrt{1-x^{2}} \frac{x}{\left(1-x^{2}\right)} y=x^{4}+2 x\) \(\frac{\mathrm{d}}{\mathrm{dx}}\left(\sqrt{1-\mathrm{x}^{2}} \mathrm{y}\right)=\mathrm{x}^{4}+2 \mathrm{x}\) \(\int d\left(\sqrt{1-x^{2}} y\right)=\int\left(x^{4}+2 x\right) d x\) \(y \sqrt{1-x^{2}}=\frac{x^{5}}{5}+\frac{2 x^{2}}{2}+C\) Using initial conditions, that is \(f(0)=0\) \(\sqrt{1-\mathrm{x}^{2}}(0)=0+0+\mathrm{C}\) Therefore, \(\mathrm{C}=0\) Therefore, equation (ii) gives \(\sqrt{1-x^{2}} y=\frac{x^{5}}{5}+x^{2}\) Hence, \(y=\frac{\frac{x^{5}}{5}+x^{2}}{\sqrt{1-x^{2}}}=\frac{x^{5}}{5 \sqrt{1-x^{2}}}+\frac{x^{2}}{\sqrt{1-x^{2}}}\) Now, \(\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x) d x=\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{5}}{5 \sqrt{1-x^{2}}} d x+\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) \(=0+2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^{2}}{\sqrt{1-x^{2}}} d x\) Since even function now solving \(=\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\mathrm{x}^{2}}{\sqrt{1-\mathrm{x}^{2}}} \mathrm{dx}\) Putting \(x=\sin \theta\). Therefore, \(d x=\cos \theta d \theta\) \(\int_{0}^{\frac{\sqrt{3}}{2}} \frac{\sin ^{2} \theta \cos \theta \mathrm{d} \theta}{\cos \theta}=\int_{0}^{\frac{\sqrt{3}}{2}}\left(\frac{1-\cos 2 \theta}{2}\right) \mathrm{d} \theta\) \(=\left[\frac{1}{2} \theta\right]_{0}^{\frac{\pi}{3}}-\frac{1}{2}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{3}}=\frac{1}{2} \times \frac{\pi}{3}-\frac{1}{4}\left[\sin \frac{2 \pi}{3}\right]\) \(=\frac{\pi}{6}-\frac{1}{4} \frac{\sqrt{3}}{2}=\frac{\pi}{6}-\frac{\sqrt{3}}{8}\) Hence, from Equation (iii), we get - \(2\left(\frac{\pi}{6}-\frac{\sqrt{3}}{8}\right)=\frac{\pi}{3}-\frac{\sqrt{3}}{4}\)
JEE Main-2022-26.07.2022
Integral Calculus
86390
The integral \(\int_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} d x\) is equal to :
