calculate-dt-cos-2t-a-cos-2t-a-

Question Number 77002 by jagoll last updated on 02/Jan/20

c a l c u l a t e \int \frac{d t}{\cos (2 t + a) \cos (2 t - a)}

Commented by turbo msup by abdo last updated on 03/Jan/20

w e h a v e

c o s (2 t + a) c o s (2 t - a)

= \frac{1}{2} (c o s (4 t) + c o s (2 a)) \Rightarrow

I = 2 \int \frac{d t}{c o s (4 t) + c o s (2 a)}

=_{4 t = u} 2 \int \frac{1}{c o s (u) + c o s (2 a)} \frac{d u}{4}

= \frac{1}{2} \int \frac{d u}{c o s (u) + c o s (2 a)}

=_{t a n (\frac{u}{2}) = x} \frac{1}{2} \int \frac{1}{\frac{1 - x^{2}}{1 + x^{2}} + c o s (2 a)} \frac{2 d x}{1 + x^{2}}

= \int \frac{d x}{1 - x^{2} + c o s (2 a) + c o s (2 a) x^{2}}

= \int \frac{d x}{(c o s (2 a) - 1) x^{2} + 1 + c o s (2 a)}

= \int \frac{d x}{1 + c o s (2 a) - (1 - c o s (2 a) x^{2}}

= \frac{1}{1 + c o s (2 a)} \int \frac{d x}{1 - \frac{1 - c o s (2 a)}{1 + c o s (2 a)} x^{2}}

= \frac{1}{1 + c o s (2 a)} \int \frac{d x}{1 - {(t a n a x)}^{2}}

=_{x t a n a = z} \frac{1}{1 + c o s (2 a)} \int \frac{d z}{t a n a (1 - z^{2})}

= \frac{1}{1 + c o s (2 a)) t a n a} \int (\frac{1}{1 - z} + \frac{1}{1 + z}) d z

= \frac{1}{t a n a (1 + c o s (2 a))} \frac{1}{2} l n ∣ \frac{1 + z}{1 - z} ∣ + c

= \frac{1}{2 t a n a (1 + c o s (2 a))} l n ∣ \frac{1 + x t a n a}{1 - x t a n a} ∣ + c

= \frac{1}{2 t a n a (1 + c o s (2 a)} l n ∣ \frac{1 + t a n a t a n (\frac{u}{2})}{1 - x t a n a t a n (\frac{u}{2})} ∣ + C

w i t h u = 4 t

Commented by mathmax by abdo last updated on 03/Jan/20

\Rightarrow I = \frac{1}{2 t a n (a) \times (2 {c o s}^{2} a)} l n ∣ \frac{1 + t a n a . t a n (2 t)}{1 - t a n a . t a n (2 t)} ∣ + C

= \frac{1}{4 \frac{s i n a}{c o s a} \times {c o s}^{2} a} l n ∣ \frac{1 + t a n (a) t a n (2 t)}{1 - t a n a . t a n (2 t)} ∣ + C

= \frac{1}{2 s i n (2 a)} l n ∣ \frac{1 + t a n a . t a n (2 t)}{1 - t a n a . t a n (2 t)} ∣ + C .

Answered by john santu last updated on 02/Jan/20

b y m a n i p u l a t e a l g e b r a

= \int \frac{s i n [(2 t + a) - (2 t - a)]}{\sin 2 a \times \cos (2 t + a) \cos (2 t - a)} d t

= \frac{1}{\sin 2 a} \int \frac{\sin (2 t + a) \cos (2 t - a) - \cos (2 t + a) \sin (2 t - a)}{\cos (2 t + a) \cos (2 t - a)} d t

= \frac{1}{\sin 2 a} {\int \tan (2 t + a) d t - \int \tan (2 t - a) d t}

= \frac{1}{\sin 2 a} {- \frac{1}{2} l n ∣ \sec (2 t + a) ∣ + \frac{1}{2} l n ∣ \sec (2 t - a) ∣} + c . ★

Commented by peter frank last updated on 02/Jan/20

g o o d

Commented by jagoll last updated on 02/Jan/20

t h a n k s y o u s i r

Commented by peter frank last updated on 02/Jan/20

p l e a s e h e l p Q n 76964

Commented by john santu last updated on 02/Jan/20

i t r y s i r

Answered by MJS last updated on 02/Jan/20

\int \frac{d t}{\cos (2 t + a) \cos (2 t - a)} =

= 2 \int \frac{d t}{\cos 4 t + \cos 2 a} =

[u = \tan 2 t \to d t = \frac{d u}{2 (t^{2} + 1)}]

= - \frac{1}{2} \int \frac{d u}{u^{2} \sin^{2} a - \cos^{2} a} =

= \frac{1}{4 \cos a} (\int \frac{d u}{u \sin a + \cos a} - \int \frac{d u}{u \sin a - \cos a}) =

= \frac{1}{4 \cos a \sin a} (\ln (u \sin a + \cos a) - \ln (u \sin a - \cos a)) =

= \frac{1}{2 \sin 2 a} \ln \frac{u \sin a + \cos a}{u \sin a - \cos a} =

= \frac{1}{2 \sin 2 a} \ln ∣ \frac{\cos (2 t - a)}{\cos (2 t + a)} ∣ + C

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