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Question Number 105504 by bemath last updated on 29/Jul/20

(1) \lim_{x \to 0} \frac{\int_{0}^{x^{2}} \sec^{2} t d t}{x \sin x} ?

(2) \underset{- π / 2}{\int^{π / 2}} \sqrt{\sec x - \cos x} d x ?

(3) I n a t r i a n g l e i f \tan A = 2 \sin 2 C

a n d 3 \cos A = 2 \sin B \sin C . f i n d C

Answered by Dwaipayan Shikari last updated on 29/Jul/20

1) \lim_{x \to 0} \frac{\int_{0}^{x^{2}} {s e c}^{2} t d t}{x s i n x} = \lim_{x \to 0} \frac{{[t a n t]}_{0}^{x^{2}}}{x^{2}} {s i n x \to x

= \lim_{x \to 0} \frac{t a n x^{2}}{x^{2}} = 1

Answered by john santu last updated on 29/Jul/20

(1) \lim_{x \to 0} \frac{2 x \sec^{2} (x^{2})}{\sin x + x \cos x} =

\lim_{x \to 0} \frac{2 x}{\sin (x) \cos^{2} (x^{2}) + x \cos x} =

\lim_{x \to 0} \frac{2}{(\frac{\sin x}{x}) \cos^{2} (x^{2}) + \cos x} = \frac{2}{1 + 1} = 1

(J S ♠ ★)

Answered by Ar Brandon last updated on 29/Jul/20

1 ∖ \lim_{x \to 0} \frac{\int_{0}^{x^{2}} \sec^{2} tdt}{xsinx} = \lim_{x \to 0} \frac{{[tanx]}_{0}^{x^{2}}}{xsinx} = \lim_{x \to 0} \frac{{tanx}^{2}}{xsinx}

= \lim_{x \to 0} \frac{x^{2}}{xsinx} = \lim_{x \to 0} \frac{x}{sinx} = \lim_{x \to 0} \frac{1}{cosx} = 1

Answered by Dwaipayan Shikari last updated on 29/Jul/20

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sqrt{1 - {c o s}^{2} x}}{\sqrt{c o s x}} = - 2 \int_{0}^{\frac{π}{2}} - \frac{s i n x}{\sqrt{c o s x}} d x = - 2 \int_{1}^{0} \frac{2 t d t}{t} = 4 \int_{0}^{1} d t = 4

c o s x = t^{2}

Commented by Dwaipayan Shikari last updated on 29/Jul/20

Hey Brandon sir , can you take a try on my question?�� Question no 105498

Commented by Ar Brandon last updated on 29/Jul/20

��OK bro, but I'm not Sir. Next 2 years I'll be thrice as old as I was e^2.3 years ago to the nearest round figure.��

Commented by Dwaipayan Shikari last updated on 29/Jul/20

��My current age is e^2.79

Commented by Dwaipayan Shikari last updated on 29/Jul/20

A high school one

Commented by Ar Brandon last updated on 29/Jul/20

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Answered by Ar Brandon last updated on 29/Jul/20

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{secx - cosx} dx = 2 \int_{0}^{\frac{π}{2}} \sqrt{\frac{1 - \cos^{2} x}{cosx}} dx = 2 \int_{0}^{\frac{π}{2}} \frac{sinx}{\sqrt{cosx}} dx

= - 2 \int_{0}^{\frac{π}{2}} \frac{d (cosx)}{\sqrt{cosx}} = - 4 {[\sqrt{cosx}]}_{0}^{\frac{π}{2}} = 4

Answered by mathmax by abdo last updated on 29/Jul/20

1) let f (x) = \frac{\int_{0}^{x^{2}} \sec^{2} tdt}{xsinx} \Rightarrow f (x) = \frac{\tan (x^{2})}{xsinx} \sim \frac{x^{2}}{x^{2}} = 1 \Rightarrow

\lim_{x \to 0} f (x) = 1

Answered by mathmax by abdo last updated on 29/Jul/20

2) I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{\frac{1}{cosx} - cosx} dx = \int_{- \frac{π}{2}}^{\frac{π}{2}} \sqrt{\frac{1 - \cos^{2} x}{cosx}} dx

= 2 \int_{0}^{\frac{π}{2}} \frac{sinx}{\sqrt{cosx}} dx =_{\sqrt{cosx} = t} - 2 \int_{1}^{0} \frac{\sqrt{1 - t^{4}}}{t} \frac{2 t}{\sqrt{1 - t^{4}}} dt = 4 \int_{0}^{1} dt = 4

(cosx = t^{2} \Rightarrow x = arcos (t^{2}))

Answered by bramlex last updated on 29/Jul/20

(3) A = 180 ° - (B + C)

\cos A = - \cos (B + C)

\to \frac{2 \sin B \sin C}{3} = - {\cos B \cos C - \sin B \sin C}

2 \sin B \sin C = - 3 \cos B \cos C - 3 \sin B \sin C

5 \sin B \sin C = - 3 \cos B \cos C

\to \tan B \tan C = - \frac{3}{5} \dots (1)

\tan A = - \tan (B + C)

\to 2 \sin 2 C = - {\frac{\tan B + \tan C}{1 - \tan B \tan C}}

\to 2 \sin 2 C = - {\frac{\tan B + \tan C}{\frac{8}{5}}}

\frac{16}{5} \sin 2 C = - \tan B - \tan C \dots (2)

Answered by Ar Brandon last updated on 29/Jul/20

3 ∖

A + B + C = π \dots \dots \dots (1)

tanA = 2 \sin 2 C \dots \dots (2)

3 cosA = 2 sinBsinC (3)

(1) in (3) \Rightarrow 3 cosA = 2 \sin (π - (A + C)) sinC

\Rightarrow 3 cosA = 2 \sin (A + B) sinC

\Rightarrow - 3 cosA = \cos (A + 2 C) - cosA

\Rightarrow - 2 cosA = \cos (A + 2 C)

From (2), tanA = 2 \sin 2 C

\Rightarrow sinA = 2 \sin 2 CcosA

\Rightarrow sinA = \sin (A + 2 C) + \sin (A - 2 C)

S t u c k e d!!!