If the equality ∫0x btcos⁡4t−asin⁡4tt2dt=asin⁡4xx−1holds for all x such that  0

∫0x btcos⁡4t−asin⁡4tt2dt=b∫0xcos4ttdt-a∫0xsin4tt2 dt=b∫0x cos⁡4ttdt−a−sin⁡4tt0x+4∫0x cos⁡4ttdt=(b−4a)∫0x cos⁡4ttdt+asin⁡4xx−4aThus (b−4a)∫0xxcos⁡4ttdt+ asin⁡4xx−4a=asin⁡4xx−1i.e. (b−4a)∫0xxcos⁡4ttdt=4a-1.Since R.H.S. is independent of x, so we must have b−4a=0 and 4a−1=0 i.e., a=1/4,b=1.