Integration-2

Calculus Level 4

If ∫ d x ( sin ⁡ x ) ( cos ⁡ x ) 2 \large \displaystyle \int{\dfrac{dx}{(\sin x)(\cos x)^2}} is equal to

a b sec ⁡ x + c d ln ⁡ tan ⁡ x e + z \dfrac{a}{b} \sec x +\dfrac{c}{d} \ln \tan \dfrac{x}{e}+z where z z is integration constant where a , b , c , d a,b,c,d are least possible natural numbers.

Then the value of a + b + c + d + e a+b+c+d+e is

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The answer is 6.

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Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Akshay Sharma
Mar 12, 2016
  • ∫ d x ( s i n x ) ( c o s x ) 2 \large \int{\frac{dx}{(sin x)(cos x)^2}}
  • ⇒ ∫ ( 1 ) d x ( s i n x ) ( c o s x ) 2 \Rightarrow\int{\frac{(1) dx}{(sin x)(cos x)^2}}
  • ⇒ \Rightarrow We can write 1 = ( s i n x ) 2 + ( c o s x ) 2 1=(sin x)^2+(cos x)^2
  • ⇒ ∫ ( s i n x ) 2 + ( c o s x ) 2 d x ( s i n x ) ( c o s x ) 2 \Rightarrow\int{\frac{(sin x)^2+(cos x)^2 dx}{(sin x)(cos x)^2}}
  • ⇒ ∫ ( s e c x ) ( t a n x ) d x + ∫ ( c o s e c x ) ( d x ) \Rightarrow\int{(sec x)(tan x) dx}+\int{(cosec x)(dx)}
  • ⇒ s e c x + l n t a n x x 2 + z \Rightarrow sec x+ln tan x\frac{x}{2}+z
  • ⇒ a = b = c = d = 1 \Rightarrow a=b=c=d=1 & e = 2 e=2
  • ⇒ a + b + c + d + e = 6 \Rightarrow a+b+c+d+e=6
2 Helpful 0 Interesting 0 Brilliant 0 Confused

In Question you have not mentioned lntanx (you write tanx only)

Yash Dev Lamba - 5 years, 3 months ago

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yup! I have edited it.

Akshay Sharma - 5 years, 3 months ago

Just for the sake of variety, the integrand can be simplified by the following ways too :

  1. Write 1 cos ⁡ 2 ( x ) \large \frac{1}{ \cos^2 (x)} as sec ⁡ 2 x = 1 + tan ⁡ 2 x \large \sec^2 x = 1 + \tan^2 x .

OR

  1. Rewrite cos ⁡ 2 x = 1 − sin ⁡ 2 x \large \cos^2 x = 1 - \sin^2 x . Rewrite numerator as sin ⁡ 2 x + 1 − sin ⁡ 2 x \large \sin^2 x + 1 - \sin^2 x .

Both ways yields the result sec ⁡ x tan ⁡ x + c o s e c x \large \sec x \tan x + \ cosec x once simplified.

Pulkit Gupta - 5 years, 3 months ago

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