I-0-pi-4-sin-4x-cos-2-x-tan-4-x-1-dx-

Question Number 87839 by jagoll last updated on 06/Apr/20

I = \underset{0}{\int^{\frac{π}{4}}} \frac{\sin 4 x}{\cos^{2} x \sqrt{\tan^{4} x + 1}} dx

Answered by redmiiuser last updated on 06/Apr/20

\sqrt{\tan^{4} x + 1}

= \sqrt{\sin^{4} x + \cos^{4} x} / \cos^{2} x

\frac{\sin 4 x}{\sqrt{\sin^{4} x + \cos^{4} x}}

= \frac{\sin 4 x}{\sqrt{\frac{3 + \cos 4 x}{4}}}

\int_{0}^{\frac{π}{4}} \frac{2 \sin 4 x}{\sqrt{3 + \cos 4 x}} d x

3 + \cos 4 x = z

d z = 4 . \sin 4 x . d x

\int_{4}^{2} \frac{2 . d z}{4 . \sqrt{z}}

= \frac{1}{2} \int_{4}^{2} \frac{d z}{\sqrt{z}}

= {[\sqrt{z}]}_{4}^{2}

= \sqrt{2} - 2

Commented by jagoll last updated on 06/Apr/20

yes sir . thank you

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