If Φ ( x ) = ∫ d x s i n 1 2 x c o s 7 2 x , then Φ ( π 4 ) − Φ ( 0 ) = Options: A . &nbsp 12 5 B . &nbsp 9 5 C . &nbsp 6 5 D . &nbsp 0

Answer: Option A : A t a n x = t ⇒ s e c 2 x d x = d t ∴ f ( x ) = ∫ d x s i n 1 2 x c o s 1 2 x . c o s 4 x = ∫ ( 1 + t a n 2 x ) s e c 2 x √ t a n x d x = ∫ ( 1 + t 2 ) √ t d t = ∫ ( t − 1 / 2 + t 3 / 2 ) d t = 2 t 1 / 2 + 2 5 t 5 / 2 = 2 √ t a n x + 2 5 ( t a n x ) 5 / 2 ∴ ϕ ( π 4 ) − ϕ ( 0 ) = 2 + 2 5 = 12 5

If Φ(x)=∫dxsin12x cos72x, then Φ(π4)−

Discussion Forum : Indefinite Integration

Question -

If $Φ (x) = \int \frac{d x}{s i n^{\frac{1}{2}} x c o s^{\frac{7}{2}} x}$ , then $Φ (\frac{π}{4}) - Φ (0) =$

Options:

A .

\frac{12}{5}

B .

\frac{9}{5}

C .

\frac{6}{5}

D . 0

Answer: Option A
:
A

t a n x = t \Rightarrow s e c^{2} x d x = d t ∴ f (x) = \int \frac{d x}{\frac{s i n^{\frac{1}{2}} x}{c o s^{\frac{1}{2}} x} . c o s^{4} x} = \int \frac{(1 + t a n^{2} x) s e c^{2} x}{\sqrt{t a n x}} d x = \int \frac{(1 + t^{2})}{\sqrt{t}} d t = \int (t^{- 1 / 2} + t^{3 / 2}) d t = 2 t^{1 / 2} + \frac{2}{5} t^{5 / 2} = 2 \sqrt{t a n x} + \frac{2}{5} (t a n x)^{5 / 2} ∴ ϕ (\frac{π}{4}) - ϕ (0) = 2 + \frac{2}{5} = \frac{12}{5}

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Discussion Forum : Indefinite Integration

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